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.

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.

Examples

Input

3
1 2
2 3


Output

125000001


Input

4
1 2
2 3
3 4


Output

375000003


Input

4
1 2
1 3
1 4


Output

250000002


Input

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


Output

570312505
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define MOD 1000000007

int N;
vector<int> G[200100];
int memo[200100];
int ans = 0;

int mod_pow(int a, int n) {
    int res = 1;
    while (n>0) {
        if (n&1) res = res*a%MOD;
        a = a*a%MOD;
        n >>= 1;
    }
    return res;
}

int dfs(int v, int pv) {
    if (memo[v]!=-1) return memo[v];
    
    int res = 1;
    int add = mod_pow(2, N-1)-1;
    
    for (auto nv : G[v]) {
        if (nv==pv) continue;
        add -= mod_pow(2, dfs(nv, v))-1;
        if (add<0) add += MOD;
        res += dfs(nv, v);
    }
    
    add -= mod_pow(2, N-res)-1;
    if (add<0) add += MOD;
    ans += add;
    ans %= MOD;
    
    return memo[v] = res;
}

signed main() {
    cin >> N;
    
    for (int i=0; i<N-1; i++) {
        int ai, bi; cin >> ai >> bi;
        G[ai-1].push_back(bi-1);
        G[bi-1].push_back(ai-1);
    }
    
    for (int i=0; i<N; i++) memo[i] = -1;
    dfs(0, -1);
    
    ans *= mod_pow(mod_pow(2, N), MOD-2);
    ans %= MOD;
    
    cout << ans << endl;
}