Problem:
Score : 600 points

Problem Statement

Given is a tree T with N vertices. The i-th edge connects Vertex A_i and B_i (1 \leq A_i,B_i \leq N).

Now, each vertex is painted black with probability 1/2 and white with probability 1/2, which is chosen independently from other vertices. Then, let S be the smallest subtree (connected subgraph) of T containing all the vertices painted black. (If no vertex is painted black, S is the empty graph.)

Let the holeyness of S be the number of white vertices contained in S. Find the expected holeyness of S.

Since the answer is a rational number, we ask you to print it \bmod 10^9+7, as described in Notes.

Notes

When you print a rational number, first write it as a fraction \frac{y}{x}, where x, y are integers, and x is not divisible by 10^9 + 7
(under the constraints of the problem, such representation is always possible).

Then, you need to print the only integer z between 0 and 10^9 + 6, inclusive, that satisfies xz \equiv y \pmod{10^9 + 7}.

Constraints

2 \leq N \leq 2 \times 10^5

1 \leq A_i,B_i \leq N

The given graph is a tree.

Input

Input is given from Standard Input in the following format:

N
A_1 B_1
:
A_{N-1} B_{N-1}

Output

Print the expected holeyness of S, \bmod 10^9+7.

Sample Input 1

3
1 2
2 3

Sample Output 1

125000001

If the vertices 1, 2, 3 are painted black, white, black, respectively, the holeyness of S is 1.

Otherwise, the holeyness is 0, so the expected holeyness is 1/8.

Since 8 \times 125000001 \equiv 1 \pmod{10^9+7}, we should print 125000001.

Sample Input 2

4
1 2
2 3
3 4

Sample Output 2

375000003

The expected holeyness is 3/8.

Since 8 \times 375000003 \equiv 3 \pmod{10^9+7}, we should print 375000003.

Sample Input 3

4
1 2
1 3
1 4

Sample Output 3

250000002

The expected holeyness is 1/4.

Sample Input 4

7
4 7
3 1
2 6
5 2
7 1
2 7

Sample Output 4

570312505

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const int inv2 = 500000004;
const int MAXN = 200005;
int n, ans, pw2[MAXN];
int fir[MAXN], to[MAXN<<1], nxt[MAXN<<1], cnt;
inline void link(int u, int v) {
	to[++cnt] = v; nxt[cnt] = fir[u]; fir[u] = cnt;
	to[++cnt] = u; nxt[cnt] = fir[v]; fir[v] = cnt;
}
int sz[MAXN];
void dfs(int u, int ff) {
	sz[u] = 1;
	int p0 = 1, p1 = 0, q0, q1, p;
	for(int i = fir[u], v; i; i = nxt[i])
		if((v=to[i]) != ff) {
			dfs(v, u); sz[u] += sz[v];
			p = pw2[sz[v]];
			q0 = 1ll * p0 * p % mod;
			q1 = (1ll * p1 * p + 1ll * p0 * (mod+1-p)) % mod;
			p0 = q0, p1 = q1;
		}
	p = pw2[n-sz[u]];
	q0 = 1ll * p0 * p % mod;
	q1 = (1ll * p1 * p + 1ll * p0 * (mod+1-p)) % mod;
	p0 = q0, p1 = q1;
	ans = (ans + (1ll+mod+mod-p0-p1) % mod * inv2 % mod) % mod;
}
int main () {
	scanf("%d", &n);
	pw2[0] = 1;
	for(int i = 1, x, y; i < n; ++i)
		scanf("%d%d", &x, &y), link(x, y), pw2[i] = 1ll * pw2[i-1] * inv2 % mod;
	pw2[n] = 1ll * pw2[n-1] * inv2 % mod;
	dfs(1, 0);
	printf("%d\n", ans);
}