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 (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
int a[200000], b[200000];
int *tree[200000], *tree_d[200000], cnt[200000], cnts[200000];
int n;
int DFS(int x, int y){//xが今いる頂点，yが前にいた頂点
    //2つの値を求める，tree[x][x_cnt] = y, tree[y][y_cnt] = xとなる
    //x_cnt, y_cnt
    int i;
    int x_cnt, y_cnt;
    for(i = 0; i < cnt[x]; i++){
        if(tree[x][i] == y){
            x_cnt = i;
            break;
        }
    }
    
    for(i = 0; i < cnt[y]; i++){
        if(tree[y][i] == x){
            y_cnt = i;
            break;
        }
    }
    
    tree_d[x][x_cnt] = 1;
    for(i = 0; i < cnt[x]; i++){
        if(tree[x][i] == y){
            continue;
        }
        tree_d[x][x_cnt] += DFS(tree[x][i], x);
    }

    tree_d[y][y_cnt] = n - tree_d[x][x_cnt];
    return tree_d[x][x_cnt];
}

int main(void){
    int i, j;
    scanf("%d", &n);
    for(i = 0; i < n; i++){
        cnt[i] = 0;
        cnts[i] = 0;
    }

    for(i = 0; i < n - 1; i++){
        scanf("%d%d", &a[i], &b[i]);
        a[i]--;
        b[i]--;
        cnt[a[i]]++;
        cnt[b[i]]++;
    }

    for(i = 0; i < n; i++){
        tree[i] = (int *)malloc(sizeof(int) * (cnt[i] + 1));
        if(tree[i] == NULL){
            return 0;
        }
    }
    for(i = 0; i < n; i++){
        tree_d[i] = (int *)malloc(sizeof(int) * (cnt[i] + 1));
        if(tree_d[i] == NULL){
            return 0;
        }
    }
    for(i = 0; i < n; i++){
        tree[a[i]][cnts[a[i]]] = b[i];
        tree[b[i]][cnts[b[i]]] = a[i];
        tree_d[a[i]][cnts[a[i]]] = -1;
        tree_d[b[i]][cnts[b[i]]] = -1;
        cnts[a[i]]++;
        cnts[b[i]]++;
    }

    DFS(a[0], b[0]);
    DFS(b[0], a[0]);



}