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

#define int long long int

const int MAX_N = 2 * 1E5 + 10, mod = 1E9 + 7;
vector<int> gr[MAX_N];
int res = 0, n;
vector<int> powers(MAX_N);

int pw(int a, int p)
{
    int cur = a, ans = 1;
    while (p > 0) {
        if (p & 1) ans = (ans * cur) % mod;
        p /= 2;
        cur = (cur * cur) % mod;
    }
    return ans;
}

int inv(int x)
{
    return pw(x, mod - 2);
}

int DFS(int v, int p)
{
    res = (res + powers[n - 1]) % mod;
    int sz = 1, num = 0;
    for (auto d : gr[v])
    {
        if (d == p) continue;
        int cur_sz = DFS(d, v);
        res = (res - powers[cur_sz]) % mod;
        sz += cur_sz;
        num++;
    }
    if (p != -1)
    {
        int cur_sz = n - sz;
        res = (res - powers[cur_sz]) % mod;
        num++;
    }
    res = (res + num - 1) % mod;
    return sz;
}

signed main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    cin >> n;
    for (int i = 0; i < n - 1; i++)
    {
        int a, b;
        cin >> a >> b;
        a--, b--;
        gr[a].push_back(b);
        gr[b].push_back(a);
    }

    powers[0] = 1;
    for (int i = 1; i <= n; i++)
        powers[i] = (powers[i - 1] * 2) % mod;

    DFS(0, -1);

    int y = (res * inv(powers[n])) % mod;
    if (y < 0) y += mod;
    cout << y << "\n";

    return 0;
}
