#include #include typedef long long int ll; using namespace std; typedef pair P; using namespace atcoder; template using min_priority_queue = priority_queue, greater>; #define USE998244353 #ifdef USE998244353 const ll MOD = 998244353; // const double PI = 3.1415926535897932; using mint = modint998244353; #else const ll MOD = 1000000007; using mint = modint1000000007; #endif const int MAX = 2000001; long long fac[MAX], finv[MAX], inv[MAX]; void COMinit() { fac[0] = fac[1] = 1; finv[0] = finv[1] = 1; inv[1] = 1; for (int i = 2; i < MAX; i++){ fac[i] = fac[i - 1] * i % MOD; inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD; finv[i] = finv[i - 1] * inv[i] % MOD; } } long long COM(int n, int k){ if (n < k) return 0; if (n < 0 || k < 0) return 0; return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD; } ll gcd(ll x, ll y) { if (y == 0) return x; else if (y > x) { return gcd (y, x); } else return gcd(x % y, y); } ll lcm(ll x, ll y) { return x / gcd(x, y) * y; } ll my_sqrt(ll x) { // ll m = 0; ll M = 3000000001; while (M - m > 1) { ll now = (M + m) / 2; if (now * now <= x) { m = now; } else { M = now; } } return m; } ll keta(ll num, ll arity) { ll ret = 0; while (num) { num /= arity; ret++; } return ret; } ll ceil(ll n, ll m) { // n > 0, m > 0 ll ret = n / m; if (n % m) ret++; return ret; } ll pow_ll(ll x, ll n) { if (n == 0) return 1; if (n % 2) { return pow_ll(x, n - 1) * x; } else { ll tmp = pow_ll(x, n / 2); return tmp * tmp; } } vector compress(vector& v) { // [3 5 5 6 1 1 10 1] -> [1 2 2 3 0 0 4 0] vector u = v; sort(u.begin(), u.end()); u.erase(unique(u.begin(),u.end()),u.end()); map mp; for (int i = 0; i < u.size(); i++) { mp[u[i]] = i; } for (int i = 0; i < v.size(); i++) { v[i] = mp[v[i]]; } return v; } vector Eratosthenes( const ll N ) { vector is_prime( N + 1 ); for( ll i = 0; i <= N; i++ ) { is_prime[ i ] = true; } vector P; for( ll i = 2; i <= N; i++ ) { if( is_prime[ i ] ) { for( ll j = 2 * i; j <= N; j += i ) { is_prime[ j ] = false; } P.emplace_back( i ); } } return P; } // return a[0]b[1]b[2]...b[n - 1] + b[0]a[1]b[2]...b[n - 1] mint g(vector a, vector b) { int n = b.size(); int id; int cnt = 0; for (int i = 0; i < n; i++) { if (b[i].val() == 0) { id = i; cnt++; } } if (cnt >= 2) return 0; if (cnt == 1) { mint ret = 1; for (int i = 0; i < n; i++) { if (i == id) { ret *= a[i]; } else { ret *= b[i]; } } return ret; } mint product_b = 1; mint ratio_sum = 0; for (int i = 0; i < n; i++) { product_b *= b[i]; ratio_sum += a[i] * b[i].inv(); } return product_b * ratio_sum; } mint sub_f(vector c) { mint sum = 0; mint sumsq = 0; for (auto x : c) { sum += x; sumsq += x * x; } mint two = 2; return (sum * sum - sumsq) * two.inv(); } // return a[0]a[1]b[2]b[3]...b[n - 1] + a[0]b[1]a[2]...b[n - 1] mint f(vector a, vector b) { int n = b.size(); int id1, id2; int cnt = 0; for (int i = 0; i < n; i++) { if (b[i].val() == 0) { if (cnt == 0) { id1 = i; } else { id2 = i; } cnt++; } } if (cnt >= 3) return 0; if (cnt == 2) { mint ret = 1; for (int i = 0; i < n; i++) { if ((i == id1) || (i == id2)) { ret *= a[i]; } else { ret *= b[i]; } } return ret; } if (cnt == 1) { vector new_a; vector new_b; for (int i = 0; i < n; i++) { if (i == id1) { ; } else { new_a.push_back(a[i]); new_b.push_back(b[i]); } } return g(new_a, new_b) * a[id1]; } mint product_b = 1; vector c; for (int i = 0; i < n; i++) { product_b *= b[i]; c.push_back(a[i] * b[i].inv()); } mint ratio_dual_sum = sub_f(c); return product_b * ratio_dual_sum; } // return b[0]b[1]b[2]b[3]...b[n - 1] mint h(vector a, vector b) { mint ret = 1; for (auto x : b) ret *= x; return ret; } void test() { vector a1 = {2, 3, 4}; vector b1 = {0, 2, 4}; cout << f(a1, b1).val() << '\n'; cout << g(a1, b1).val() << '\n'; } mint solve(int cur, int par, bool have_par, vector>& edge) { vector a; vector b; for (auto nex : edge[cur]) { if (nex == par) continue; a.push_back(solve(nex, cur, true, edge)); b.push_back(solve(nex, cur, false, edge)); } int n = a.size(); // if (n == 0) { // if (have_par) { // return 1; // } // else { // return 0; // } // } // if (n == 1) { // if (have_par) { // return a[0] + b[0]; // } // else { // return a[0]; // } // } // 子の数が2以上だから成り立つこと if (have_par) { return g(a, b) + h(a, b); } else { return f(a, b) + g(a, b); } } int main() { int n; cin >> n; vector> edge(n); for (int i = 1; i <= n - 1; i++) { int u, v; cin >> u >> v; u--; v--; edge[u].push_back(v); edge[v].push_back(u); } cout << solve(0, -1, false, edge).val() << '\n'; return 0; }