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 i64 = long long;

i64 gcd(i64 a, i64 b) { return a ? gcd(b % a, a) : b; }

const int MOD = 1000000007;

i64 pow64(i64 a, i64 b) {
    a %= MOD;
    i64 result = 1;
    while(b) {
        if(b & 1) result = result * a % MOD;
        a = a * a % MOD;
        b >>= 1;
    }
    return result;
}
i64 inv64(i64 x) { return pow64(x, MOD - 2); }

const int MAXN = 201000;

int n;
std::vector<int> edges[MAXN];
int size[MAXN];
std::vector<int> vec[MAXN];

int dfs(int u, int p) {
    int result = 1;
    for(int v : edges[u]) {
        if(v == p) continue;
        int x = dfs(v, u);
        result += x;
        vec[u].push_back(x);
    }
    if(p != -1) vec[u].push_back(n - result);
    return size[u] = result;
}

int main() {
    std::scanf("%d", &n);
    for(int i = 0; i < n - 1; ++i) {
        int u, v;
        std::scanf("%d%d", &u, &v);
        --u, --v;
        edges[u].emplace_back(v);
        edges[v].emplace_back(u);
    }
    dfs(0, -1);
    /*for(int i = 0; i < n; ++i) {
        std::cout << i << " ->";
        for(int x : vec[i])
            std::cout << ' ' << x;
        std::cout << std::endl;
    }*/
    i64 result = 0;
    for(int i = 0; i < n; ++i) {
        for(int x : vec[i])
            result = (result + (pow64(2, x) - 1 + MOD) % MOD) % MOD;
        result += 1;
    }
    result = ((n * pow64(2, n - 1) - result) % MOD + MOD) % MOD;
    result = result * pow64(inv64(2), n) % MOD;
    std::printf("%lld\n", result);
}