func_code_string: #include <bits/stdc++.h>   #pragma GCC optimize( Ofast,unroll-loops )   #pragma loop - opt(on)   #pragma GCC target( sse,sse2,sse3,ssse3,sse4,avx,avx2,fma,tune=native )   using namespace std;   const long long mxN = 1e5 + 5;   long long n, m, Q;   bitset<mxN> vis, in, incy;   long long dsu[mxN], lca[mxN], eu[mxN], ev[mxN], tag[mxN], sum[mxN];   vector<long long> g[mxN], q[mxN];   long long get(long long x) { return x == dsu[x] ? x : dsu[x] = get(dsu[x]); }   void tarjan(long long u, long long p = -1) {     dsu[u] = u;     in[u] = 1;     for (long long v : g[u])       if (!in[v]) tarjan(v, u);     for (long long i : q[u]) {       long long v = u ^ eu[i] ^ ev[i];       if (vis[v]) {         lca[i] = get(v);       }       if (v == u) lca[v] = u;     }     vis[u] = 1;     dsu[u] = p;   }   vector<long long> stk;   void dfs(long long u, long long p = -1) {     vis[u] = in[u] = 1;     stk.emplace_back(u);     for (long long v : g[u]) {       if (v == p) continue;       if (in[v]) {         tag[v]++;         incy[v] = 1;         while (!stk.empty() and stk.back() != v) {           incy[stk.back()] = 1, stk.pop_back();         }         stk.pop_back();       }       if (vis[v]) continue;       dfs(v, u);     }     if (!stk.empty() and stk.back() == u) stk.pop_back();     in[u] = 0;   }   void add(long long u) {     vis[u] = 1;     sum[u] += tag[u];     for (long long v : g[u]) {       if (vis[v]) continue;       sum[v] += sum[u];       add(v);     }   }   signed main() {     ios_base::sync_with_stdio(0);     cin.tie(0);     ;     cin >> n >> m;     while (m--) {       long long a, b;       cin >> a >> b;       g[a].emplace_back(b);       g[b].emplace_back(a);     }     dfs(1, 0);     vis = 0;     add(1);     cin >> Q;     for (long long i = 0; i < Q; i++) {       cin >> eu[i] >> ev[i];       q[eu[i]].emplace_back(i);       q[ev[i]].emplace_back(i);     }     vis = 0;     in = 0;     tarjan(1, 0);     vector<long long> p2(n + 1);     p2[0] = 1;     for (long long i = 1; i <= n; i++) p2[i] = (p2[i - 1] << 1) % 1000000007;     for (long long i = 0; i < Q; i++) {       cout << p2[sum[eu[i]] + sum[ev[i]] - 2 * sum[lca[i]] + incy[lca[i]]]            <<   n ;     }   }   