#include #include #include #include #include #include #include // string, to_string, stoi #include // vector #include // min, max, swap, sort, reverse, lower_bound, upper_bound #include // pair, make_pair #include // tuple, make_tuple #include // int64_t, int*_t #include // printf #include // map #include // queue, priority_queue #include // set #include // stack #include // deque #include // unordered_map #include // unordered_set #include // bitset #include // isupper, islower, isdigit, toupper, tolower using namespace std; using ll = long long; #define all(A) A.begin(),A.end() #define rep(i, n) for (long long i = 0; i < (long long)(n); i++) ll Max(ll(a), ll(b), ll(c)) { return max(max(a, b), c); } ll Min(ll(a), ll(b), ll(c)) { return min(min(a, b), c); } struct UnionFind { vector par; // par[i]:iの親の番号 (例) par[3] = 2 : 3の親が2 UnionFind(int N) : par(N) { //最初は全てが根であるとして初期化 for (int i = 0; i < N; i++) par[i] = i; } int root(int x) { // データxが属する木の根を再帰で得る:root(x) = {xの木の根} if (par[x] == x) return x; return par[x] = root(par[x]); } void unite(int x, int y) { // xとyの木を併合 int rx = root(x); //xの根をrx int ry = root(y); //yの根をry if (rx == ry) return; //xとyの根が同じ(=同じ木にある)時はそのまま par[rx] = ry; //xとyの根が同じでない(=同じ木にない)時:xの根rxをyの根ryにつける } bool same(int x, int y) { // 2つのデータx, yが属する木が同じならtrueを返す int rx = root(x); int ry = root(y); return rx == ry; } }; ll Q[4000002]; ll modPow(long long a, long long n, long long p) { if (n == 0) return 1; // 0乗にも対応する場合 if (n == 1) return a % p; if (n % 2 == 1) return (a * modPow(a, n - 1, p)) % p; long long t = modPow(a, n / 2, p); return (t * t) % p; } ll Xor(ll a, ll b) { ll A = a, B = b; ll C = 0; ll k = 1; while (A > 0 || B > 0) { if (A % 2 != B % 2) { C += k; } k *= 2; A /= 2; B /= 2; } return C; } vector fact, factinv, inv; ll mod = 1e9 + 7; void prenCkModp(ll n) { fact.resize(n + 5); factinv.resize(n + 5); inv.resize(n + 5); fact.at(0) = fact.at(1) = 1; factinv.at(0) = factinv.at(1) = 1; inv.at(1) = 1; for (ll i = 2; i < n + 5; i++) { fact.at(i) = (fact.at(i - 1) * i) % mod; inv.at(i) = mod - (inv.at(mod % i) * (mod / i)) % mod; factinv.at(i) = (factinv.at(i - 1) * inv.at(i)) % mod; } } ll nCk(ll n, ll k) { return fact.at(n) * (factinv.at(k) * factinv.at(n - k) % mod) % mod; } ll dx[4] = { -1,0,1,0 }; ll dy[4] = { 0,-1,0,1 }; ll stll(string S, ll mod) { ll n = 0; ll k = 1; rep(i, S.size()) { n += (k * ll(S[S.size() - i - 1] - '0')); k *= 10; k %= mod; n %= mod; } return n; } ll gcd(ll(a), ll(b)) { ll c = a; while (a % b != 0) { c = a % b; a = b; b = c; } return b; } vector> AN; vector> se; vector> A; ll H, W; ll pa(ll x, ll y) { if (se[x][y])return AN[x][y]; else { ll m = 1; rep(i, 4) { ll tx = x + dx[i], ty = y + dy[i]; if (tx < 0 || tx >= H || ty < 0 || ty >= W)continue; if (A[tx][ty] > A[x][y])m += pa(tx, ty); m %= mod; } AN[x][y] = m%mod; se[x][y] = true; return AN[x][y]; } } int main() { ll A; cin >> A; while (A % 2 == 0) { A /= 2; } if (A != 1) { cout << "YES" << endl; } else { cout << "NO" << endl; } }