#ifndef HIDDEN_IN_VS // 折りたたみ用 // 警告の抑制 #define _CRT_SECURE_NO_WARNINGS // ライブラリの読み込み #include using namespace std; // 型名の短縮 using ll = long long; using ull = unsigned long long; // -2^63 ~ 2^63 = 9e18(int は -2^31 ~ 2^31 = 2e9) using pii = pair; using pll = pair; using pil = pair; using pli = pair; using vi = vector; using vvi = vector; using vvvi = vector; using vvvvi = vector; using vl = vector; using vvl = vector; using vvvl = vector; using vvvvl = vector; using vb = vector; using vvb = vector; using vvvb = vector; using vc = vector; using vvc = vector; using vvvc = vector; using vd = vector; using vvd = vector; using vvvd = vector; template using priority_queue_rev = priority_queue, greater>; using Graph = vvi; // 定数の定義 const double PI = acos(-1); int DX[4] = { 1, 0, -1, 0 }; // 4 近傍(下,右,上,左) int DY[4] = { 0, 1, 0, -1 }; int INF = 1001001001; ll INFL = 4004004003094073385LL; // (int)INFL = INF, (int)(-INFL) = -INF; // 入出力高速化 struct fast_io { fast_io() { cin.tie(nullptr); ios::sync_with_stdio(false); cout << fixed << setprecision(18); } } fastIOtmp; // 汎用マクロの定義 #define all(a) (a).begin(), (a).end() #define sz(x) ((int)(x).size()) #define lbpos(a, x) (int)distance((a).begin(), std::lower_bound(all(a), (x))) #define ubpos(a, x) (int)distance((a).begin(), std::upper_bound(all(a), (x))) #define Yes(b) {cout << ((b) ? "Yes\n" : "No\n");} #define rep(i, n) for(int i = 0, i##_len = int(n); i < i##_len; ++i) // 0 から n-1 まで昇順 #define repi(i, s, t) for(int i = int(s), i##_end = int(t); i <= i##_end; ++i) // s から t まで昇順 #define repir(i, s, t) for(int i = int(s), i##_end = int(t); i >= i##_end; --i) // s から t まで降順 #define repe(v, a) for(const auto& v : (a)) // a の全要素(変更不可能) #define repea(v, a) for(auto& v : (a)) // a の全要素(変更可能) #define repb(set, d) for(int set = 0, set##_ub = 1 << int(d); set < set##_ub; ++set) // d ビット全探索(昇順) #define repis(i, set) for(int i = lsb(set), bset##i = set; i < 32; bset##i -= 1 << i, i = lsb(bset##i)) // set の全要素(昇順) #define repp(a) sort(all(a)); for(bool a##_perm = true; a##_perm; a##_perm = next_permutation(all(a))) // a の順列全て(昇順) #define uniq(a) {sort(all(a)); (a).erase(unique(all(a)), (a).end());} // 重複除去 #define EXIT(a) {cout << (a) << endl; exit(0);} // 強制終了 #define inQ(x, y, u, l, d, r) ((u) <= (x) && (l) <= (y) && (x) < (d) && (y) < (r)) // 半開矩形内判定 // 汎用関数の定義 template inline ll powi(T n, int k) { ll v = 1; rep(i, k) v *= n; return v; } template inline bool chmax(T& M, const T& x) { if (M < x) { M = x; return true; } return false; } // 最大値を更新(更新されたら true を返す) template inline bool chmin(T& m, const T& x) { if (m > x) { m = x; return true; } return false; } // 最小値を更新(更新されたら true を返す) template inline T getb(T set, int i) { return (set >> i) & T(1); } template inline T smod(T n, T m) { n %= m; if (n < 0) n += m; return n; } // 非負mod // 演算子オーバーロード template inline istream& operator>>(istream& is, pair& p) { is >> p.first >> p.second; return is; } template inline istream& operator>>(istream& is, vector& v) { repea(x, v) is >> x; return is; } template inline vector& operator--(vector& v) { repea(x, v) --x; return v; } template inline vector& operator++(vector& v) { repea(x, v) ++x; return v; } #endif // 折りたたみ用 #if __has_include() #include using namespace atcoder; #ifdef _MSC_VER #include "localACL.hpp" #endif using mint = modint998244353; //using mint = static_modint<1000000007>; //using mint = modint; // mint::set_mod(m); string mint_to_frac(mint x, int v_max = 31595) { repi(dnm, 1, v_max) { int num = (x * dnm).val(); if (num == 0) { return "0"; } if (num <= v_max) { if (dnm == 1) return to_string(num); return to_string(num) + "/" + to_string(dnm); } if (mint::mod() - num <= v_max) { if (dnm == 1) return "-" + to_string(mint::mod() - num); return "-" + to_string(mint::mod() - num) + "/" + to_string(dnm); } } return to_string(x.val()); } namespace atcoder { inline istream& operator>>(istream& is, mint& x) { ll x_; is >> x_; x = x_; return is; } #ifdef _MSC_VER inline ostream& operator<<(ostream& os, const mint& x) { os << mint_to_frac(x); return os; } #else inline ostream& operator<<(ostream& os, const mint& x) { os << x.val(); return os; } #endif } using vm = vector; using vvm = vector; using vvvm = vector; using vvvvm = vector; using pim = pair; #endif #ifdef _MSC_VER // 手元環境(Visual Studio) #include "local.hpp" #else // 提出用(gcc) inline int popcount(int n) { return __builtin_popcount(n); } inline int popcount(ll n) { return __builtin_popcountll(n); } inline int lsb(int n) { return n != 0 ? __builtin_ctz(n) : 32; } inline int lsb(ll n) { return n != 0 ? __builtin_ctzll(n) : 64; } inline int msb(int n) { return n != 0 ? (31 - __builtin_clz(n)) : -1; } inline int msb(ll n) { return n != 0 ? (63 - __builtin_clzll(n)) : -1; } #define dump(...) #define dumpel(...) #define dump_list(v) #define dump_mat(v) #define input_from_file(f) #define output_to_file(f) #define Assert(b) { if (!(b)) { vc MLE(1<<30); EXIT(MLE.back()); } } // RE の代わりに MLE を出す #endif //【アダマール変換】: O(2^n n) /* * a[0..2^n) を * A[set] = Σset2 (-1)^popcount(set ∩ set2) a[set2] * なる A[0..2^n) に上書きする. */ template void hadamard(vector& a) { // verify : https://judge.yosupo.jp/problem/bitwise_xor_convolution // 具体例: // A[0] = a[0] + a[1] + a[2] + a[3] + a[4] + a[5] + a[6] + a[7] + ... // A[1] = a[0] - a[1] + a[2] - a[3] + a[4] - a[5] + a[6] - a[7] + ... // A[2] = a[0] + a[1] - a[2] - a[3] + a[4] + a[5] - a[6] - a[7] + ... // A[3] = a[0] - a[1] - a[2] + a[3] + a[4] - a[5] - a[6] + a[7] + ... // A[4] = a[0] + a[1] + a[2] + a[3] - a[4] - a[5] - a[6] - a[7] + ... // A[5] = a[0] - a[1] + a[2] - a[3] - a[4] + a[5] - a[6] + a[7] + ... // A[6] = a[0] + a[1] - a[2] - a[3] - a[4] - a[5] + a[6] + a[7] + ... // A[7] = a[0] - a[1] - a[2] + a[3] - a[4] + a[5] + a[6] - a[7] + ... int n = msb(sz(a)); rep(i, n) repb(set, n) { if (!(set & (1 << i))) { T x = a[set]; T y = a[set | (1 << i)]; a[set] = x + y; a[set + (1 << i)] = x - y; } } } //【逆アダマール変換(mint)】: O(2^n n + log(mod)) /* * A[0..2^n) を * A[set] = Σset2 (-1)^popcount(set ∩ set2) a[set2] * なる a[0..2^n) に上書きする. * * 制約:mint の法は 2 の倍数でない * * 利用:【アダマール変換】 */ void hadamard_inv(vm& A) { // verify : https://atcoder.jp/contests/abc265/tasks/abc265_h hadamard(A); // まとめて商をとらないと log(mod) 倍遅くなる. mint inv = mint(sz(A)).inv(); rep(i, sz(A)) A[i] *= inv; } vm TLE(int n, int m, vi l, vi r) { vm f(1LL << n, 1); rep(j, m) { dump("--- j:", j, "---"); dump("w:", r[j] - l[j]); vm g(1LL << n); repi(i, l[j], r[j] - 1) g[i]++; dump(g); hadamard(g); dump(g); rep(i, 1 << n) f[i] *= g[i]; } dump("f:"); dump(f); hadamard_inv(f); return f; } void zikken() { int n = 4; int N = 1 << n; auto dump2 = [&](vm a) { rep(b, n) { rep(i, N) if (lsb(i) == b) cout << right << setw(2) << a[i] << " "; cout << "|\n"[b == n - 1]; } }; repi(k, 0, N) { // dump("--- k:", k, "---"); vm g(N); rep(i, k) g[i] = 1; // dump2(g); hadamard(g); dump2(g); } exit(0); } /* r\i 1 3 5 7 9 11 13 15 | 2 6 10 14 | 4 12 | 8 0 0 0 0 0 0 0 0 0 | 0 0 0 0 | 0 0 | 0 1 1 1 1 1 1 1 1 1 | 1 1 1 1 | 1 1 | 1 2 0 0 0 0 0 0 0 0 | 2 2 2 2 | 2 2 | 2 3 1 -1 1 -1 1 -1 1 -1 | 1 1 1 1 | 3 3 | 3 4 0 0 0 0 0 0 0 0 | 0 0 0 0 | 4 4 | 4 5 1 1 -1 -1 1 1 -1 -1 | 1 -1 1 -1 | 3 3 | 5 6 0 0 0 0 0 0 0 0 | 2 -2 2 -2 | 2 2 | 6 7 1 -1 -1 1 1 -1 -1 1 | 1 -1 1 -1 | 1 1 | 7 8 0 0 0 0 0 0 0 0 | 0 0 0 0 | 0 0 | 8 9 1 1 1 1 -1 -1 -1 -1 | 1 1 -1 -1 | 1 -1 | 7 10 0 0 0 0 0 0 0 0 | 2 2 -2 -2 | 2 -2 | 6 11 1 -1 1 -1 -1 1 -1 1 | 1 1 -1 -1 | 3 -3 | 5 12 0 0 0 0 0 0 0 0 | 0 0 0 0 | 4 -4 | 4 13 1 1 -1 -1 -1 -1 1 1 | 1 -1 -1 1 | 3 -3 | 3 14 0 0 0 0 0 0 0 0 | 2 -2 -2 2 | 2 -2 | 2 15 1 -1 -1 1 -1 1 1 -1 | 1 -1 -1 1 | 1 -1 | 1 16 0 0 0 0 0 0 0 0 | 0 0 0 0 | 0 0 | 0 A_r[i] = Σj∈[0..r) (-1)^popcount(i ∩ j) */ void zikken2() { int n = 4; int N = 1 << n; cout << "r\\i"; rep(s, n) { rep(i, N) { if (lsb(i) == s) { cout << right << setw(2) << i << " "; } } cout << "|\n"[s == n - 1]; } repi(r, 0, N) { cout << right << setw(2) << r << " "; rep(s, n) { rep(i, N) { if (lsb(i) == s) { int t = ((i >> s) - 1) / 2; int q = r / (1 << (s + 1)); int m = r % (1 << (s + 1)); int val = (popcount(t & q) & 1 ? -1 : 1) * min(m, (1 << (s + 1)) - m); cout << right << setw(2) << val << " "; } } cout << "|\n"[s == n - 1]; } } exit(0); } /* r\i 1 3 5 7 9 11 13 15 | 2 6 10 14 | 4 12 | 8 0 0 0 0 0 0 0 0 0 | 0 0 0 0 | 0 0 | 0 1 1 1 1 1 1 1 1 1 | 1 1 1 1 | 1 1 | 1 2 0 0 0 0 0 0 0 0 | 2 2 2 2 | 2 2 | 2 3 1 -1 1 -1 1 -1 1 -1 | 1 1 1 1 | 3 3 | 3 4 0 0 0 0 0 0 0 0 | 0 0 0 0 | 4 4 | 4 5 1 1 -1 -1 1 1 -1 -1 | 1 -1 1 -1 | 3 3 | 5 6 0 0 0 0 0 0 0 0 | 2 -2 2 -2 | 2 2 | 6 7 1 -1 -1 1 1 -1 -1 1 | 1 -1 1 -1 | 1 1 | 7 8 0 0 0 0 0 0 0 0 | 0 0 0 0 | 0 0 | 8 9 1 1 1 1 -1 -1 -1 -1 | 1 1 -1 -1 | 1 -1 | 7 10 0 0 0 0 0 0 0 0 | 2 2 -2 -2 | 2 -2 | 6 11 1 -1 1 -1 -1 1 -1 1 | 1 1 -1 -1 | 3 -3 | 5 12 0 0 0 0 0 0 0 0 | 0 0 0 0 | 4 -4 | 4 13 1 1 -1 -1 -1 -1 1 1 | 1 -1 -1 1 | 3 -3 | 3 14 0 0 0 0 0 0 0 0 | 2 -2 -2 2 | 2 -2 | 2 15 1 -1 -1 1 -1 1 1 -1 | 1 -1 -1 1 | 1 -1 | 1 16 0 0 0 0 0 0 0 0 | 0 0 0 0 | 0 0 | 0 A_r[i] = Σj∈[0..r) (-1)^popcount(i ∩ j) i = (2t+1) 2^s, r = q 2^(s+1) + m のとき A_r[i] = (-1)^popcount(t ∩ q) min(m, 2^(s+1)-m) */ vm TLE2(int n, int m, vi l, vi r) { int N = 1 << n; vm res(N, 1); vm sub(N, 1); rep(j, m) res[0] *= r[j] - l[j]; rep(s, n) { int Ns = N >> (s + 1); vi ql(m), A(m), qr(m), B(m); rep(j, m) { qr[j] = r[j] / (1 << (s + 1)); ql[j] = l[j] / (1 << (s + 1)); int mr = r[j] % (1 << (s + 1)); A[j] = min(mr, (1 << (s + 1)) - mr); int ml = l[j] % (1 << (s + 1)); B[j] = min(ml, (1 << (s + 1)) - ml); } rep(t, Ns) { int i = (2 * t + 1) << s; rep(j, m) { int sgnA = popcount(t & qr[j]) & 1 ? -1 : 1; int sgnB = popcount(t & ql[j]) & 1 ? -1 : 1; res[i] *= sgnA * A[j] - sgnB * B[j]; sub[i] *= sgnB; } } } dump(res); dump(sub); hadamard_inv(res); return res; } /* f[i] = Π(A_r[i] - A_l[i]) i = (2t+1) 2^s, r = q 2^(s+1) + m のとき f[i] = Π((-1)^popcount(t ∩ qr) min(mr, 2^(s+1)-mr) - (-1)^popcount(t ∩ ql) min(ml, 2^(s+1)-ml)) */ vm TLE3(int n, int m, vi l, vi r) { int N = 1 << n; vm res(N, 1); rep(j, m) res[0] *= r[j] - l[j]; rep(s, n) { int Ns = N >> (s + 1); vi cnt(n - s); rep(j, m) { int q = r[j] >> (s + 1); repis(b, q) cnt[b]++; } rep(t, Ns) { int i = (2 * t + 1) << s; int sum = 0; repis(b, t) sum += cnt[b]; res[i] *= sum & 1 ? -1 : 1; } vi q(m), A(m), B(m); rep(j, m) { q[j] = (r[j] ^ l[j]) / (1 << (s + 1)); int mr = r[j] % (1 << (s + 1)); A[j] = min(mr, (1 << (s + 1)) - mr); int ml = l[j] % (1 << (s + 1)); B[j] = min(ml, (1 << (s + 1)) - ml); } rep(t, Ns) { int i = (2 * t + 1) << s; rep(j, m) { int sgn = popcount(t & q[j]) & 1 ? -1 : 1; res[i] *= A[j] - sgn * B[j]; } } } dump(res); hadamard_inv(res); return res; } /* f[i] = Π(A_r[i] - A_l[i]) i = (2t+1) 2^s, r = q 2^(s+1) + m のとき f[i] = Π(-1)^popcount(t ∩ qr) Π(min(mr, 2^(s+1)-mr) - (-1)^popcount(t ∩ (qr XOR ql)) min(ml, 2^(s+1)-ml)) = Π(-1)^popcount(t ∩ qr) Π_j (A_j - (-1)^popcount(t ∩ q_j) B_j) */ //【アダマール変換】O(2^n n)(の改変) /* * a[0..2^n) を * A[set] = Σset2 (-1)^popcount(set ∩ set2) a[set2] * なる A[0..2^n) に上書きする. */ void hadamard2(vector>& a) { int n = msb(sz(a)); rep(i, n) repb(set, n) { if (!(set & (1 << i))) { auto [xu, xv] = a[set]; auto [yu, yv] = a[set | (1 << i)]; // (log xu - log xv) + (log yu - log yv) // = log (xu yu) - log (xv yv) a[set] = { xu * yu, xv * yv }; // (log xu - log xv) - (log yu - log yv) // = log (xu yv) - log (xv yu) a[set + (1 << i)] = { xu * yv, xv * yu }; } } } vm solve(int n, int m, vi l, vi r) { int N = 1 << n; vm res(N, 1); rep(j, m) res[0] *= r[j] - l[j]; rep(s, n) { int Ns = N >> (s + 1); vi cnt(n - s); rep(j, m) { int q = r[j] >> (s + 1); repis(b, q) cnt[b]++; } rep(t, Ns) { int i = (2 * t + 1) << s; int sum = 0; repis(b, t) sum += cnt[b]; res[i] *= sum & 1 ? -1 : 1; } vector> CD(Ns, { 1, 1 }); rep(j, m) { int q = (r[j] ^ l[j]) / (1 << (s + 1)); int mr = r[j] % (1 << (s + 1)); int A = min(mr, (1 << (s + 1)) - mr); int ml = l[j] % (1 << (s + 1)); int B = min(ml, (1 << (s + 1)) - ml); CD[q % Ns].first *= A - B; CD[q % Ns].second *= A + B; } hadamard2(CD); rep(t, Ns) { int i = (2 * t + 1) << s; res[i] *= CD[t].first; } } dump(res); hadamard_inv(res); return res; } int main() { // input_from_file("input.txt"); // output_to_file("output.txt"); // zikken2(); int n, m; cin >> n >> m; vi l(m), r(m); rep(j, m) cin >> l[j] >> r[j]; ++r; dump(TLE(n, m, l, r)); dump("-----"); dump(TLE2(n, m, l, r)); dump("-----"); dump(TLE3(n, m, l, r)); dump("-----"); auto res = solve(n, m, l, r); repb(set, n) cout << res[set] << "\n"; }