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 <vector>
#include <iostream>
using namespace std;

#define M 1000000007

long long modexp(int x, long long e, int m) {
    long long ans = 1, p = x % m;
    while (e > 0) {
        if (e % 2 != 0) ans = (ans * p) % m;
        p = (p * p) % m;
        e >>= 1;
    }
    return ans;
}

long long f[200100], g[200100];
int s[200100];
vector<int> adj[200100];

void dfs(int i, int p) {
    int ssum = 0;
    for (int j : adj[i]) {
        if (j != p) { dfs(j, i); ssum += s[j]; }
    }
    long long fval = 0, gval = 0;
    fval = gval = modexp(2, ssum, M) - 1;
    for (int j : adj[i]) {
        fval += f[j] - modexp(2, s[j], M) + 1
            + g[j] * (modexp(2, ssum - s[j] + 1, M) - 1);
        gval += g[j] * modexp(2, ssum - s[j] + 1, M);
        fval %= M; gval %= M;
    }
    f[i] = fval;
    g[i] = gval;
    s[i] = 1 + ssum;
}

int main() {
    int n; cin >> n;
    for (int i = 1; i < n; i++) {
        int a, b; cin >> a >> b;
        a--;b--;
        adj[a].push_back(b);
        adj[b].push_back(a);
    }
    dfs(0, -1);
    cout << (f[0] * modexp((M+1)/2, s[0], M) % M + M) % M << endl;
}
