传送门

首先，所有连通块的个数的期望再减去每个点孤立的概率就是答案。
设 \(d_i\) 表示 \(i\) 的度数，那么每个点孤立的概率为 \(\frac{1}{2^{d_i}}\)
考虑计算所有连通块的个数的期望
对于一棵树来说，每次删除一条边会使得连通块的个数 \(+1\)，概率为 \(\frac{1}{2}\)，那么 \(n-1\) 条边的期望就是 \(1+\frac{n-1}{2}\)
对于仙人掌来说，如果这次删的是环上第一个被删除的边，那么不会贡献答案，所以要减去在一个环上至少删除了一条边的概率，设长度为 \(len\)，就要减去 \(1-\frac{1}{2^{len}}\)
所以只要求出每个环就好了。

# include <bits/stdc++.h>
using namespace std;
typedef long long ll;namespace IO {const int maxn(1 << 21 | 1);char ibuf[maxn], *iS, *iT, c;int f;inline char Getc() {return (iS == iT ? (iT = (iS = ibuf) + fread(ibuf, 1, maxn, stdin), (iS == iT ? EOF : *iS++)) : *iS++);}template <class Int> inline void In(Int &x) {for (f = 1, c = Getc(); c < '0' || c > '9'; c = Getc()) f = c == '-' ? -1 : 1;for (x = 0; c <= '9' && c >= '0'; c = Getc()) x = (x << 3) + (x << 1) + (c ^ 48);x *= f;}
}using IO :: In;const int maxn(1e6 + 5);
const int mod(1e9 + 7);inline void Inc(int &x, const int y) {x = x + y >= mod ? x + y - mod : x + y;
}inline void Dec(int &x, const int y) {x = x - y < 0 ? x - y + mod : x - y;
}inline int Add(int x, const int y) {return x + y >= mod ? x + y - mod : x + y;
}inline int Sub(int x, const int y) {return x - y < 0 ? x - y + mod : x - y;
}inline int Pow(ll x, int y) {ll ret = 1;for (; y; y >>= 1, x = x * x % mod)if (y & 1) ret = ret * x % mod;return ret;
}struct Edge {  int to, next;  };int n, m, d[maxn], first[maxn], cnt, inv2[maxn << 1], fa[maxn], ans, vis[maxn], deep[maxn];
Edge edge[maxn << 2];inline void AddEdge(int u, int v) {edge[cnt] = (Edge){v, first[u]}, first[u] = cnt++, ++d[u];edge[cnt] = (Edge){u, first[v]}, first[v] = cnt++, ++d[v];
}void Dfs(int u, int ff) {int e, v, len, cur;vis[u] = 1;for (e = first[u]; ~e; e = edge[e].next)if ((v = edge[e].to) ^ ff) {if (!vis[v]) deep[v] = deep[u] + 1, fa[v] = u, Dfs(v, u);else if (deep[v] < deep[u]) {for (len = 1, cur = u; cur ^ v; cur = fa[cur]) ++len;Dec(ans, (ll)Sub(1, inv2[len]) % mod);}}
}int main() {int i, u, v;memset(first, -1, sizeof(first));In(n), In(m);inv2[0] = 1, inv2[1] = (mod + 1) >> 1;for (v = m + m, i = 2; i <= v; ++i) inv2[i] = (ll)inv2[i - 1] * inv2[1] % mod;for (i = 1; i <= m; ++i) In(u), In(v), AddEdge(u, v);ans = Add(1, (ll)m * inv2[1] % mod), Dfs(1, 0);for (i = 1; i <= n; ++i) Dec(ans, inv2[d[i]]);ans = (ll)ans * Pow(2, m) % mod, printf("%d\n", ans);return 0;
}