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 ll long long

const int N = 2e5 + 5, mod = 1e9 + 7;
vector<int> g[N];
int sz[N];
int n;

int fp(int a, int p) {
    if (!p) return 1;
    int t = fp(a, p >> 1);
    t = (t * 1LL * t) % mod;
    if (p & 1) t = (t * 1LL * a) % mod;
    return t;
}

int ans = 0;

void dfs(int node, int par) {
    sz[node] = 1;
    for (auto child:g[node]) {
        if (child == par)continue;
        dfs(child, node);
        sz[node] += sz[child];
    }
    int prop = fp(fp(2, n - 1), mod - 2);
    for (auto child:g[node]) {
        if (child == par)continue;
        int x = sz[child];
        //2^x - 1 / 2 ^ n
        prop = (prop + ((fp(2, x) - 1 + mod) % mod * 1LL * fp(fp(2, n - 1), mod - 2)) % mod) % mod;
    }
    int x = n - sz[node];
    prop = (prop + ((fp(2, x) - 1 + mod) % mod * 1LL * fp(fp(2, n - 1), mod - 2)) % mod) % mod;
    prop = (1 - prop + mod) % mod;
    prop = (prop * 1LL * fp(2, mod - 2)) % mod;
    ans = (ans + prop) % mod;
}

int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cin >> n;
    for (int i = 0; i + 1 < n; i++) {
        int u, v;
        cin >> u >> v;
        g[u].push_back(v);
        g[v].push_back(u);
    }
    dfs(1, 1);
    cout << ans << endl;
    return 0;
}