#include <bits/stdc++.h>
using namespace std;

const long mod = 1000000007;
const long inv2 = 500000004;
int n;
vector<int> g[200000];
long sz[200000];

int rec(int from, int prev) {
  int sum = 1;
  for (int to : g[from]) {
    if (to == prev) continue;
    sum += rec(to, from);
  }
  sz[from] = sum;
  return sum;
}

long p[200001];
int main() {
  cin >> n;
  for (int i = 1; i < n; ++i) {
    int a, b;
    cin >> a >> b;
    --a; --b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  
  p[0] = 1;
  for (int i = 0; i < n; ++i) {
    p[i + 1] = p[i] * inv2 % mod;
  }
  
  rec(0, -1);
  long sum = 0;
  for (int i = 1; i < n; ++i) {
    long a = (1 + mod - p[sz[i]]) % mod * (1 + mod - p[n - sz[i]]) % mod;
    sum = (sum + a) % mod;
  }
  sum = (sum + mod - n * inv2 % mod) % mod;
  sum = (sum + mod - p[n] + 1) % mod;
  cout << sum << endl;
}
### Prompt

Your challenge is to write a CPP solution to the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;

const long mod = 1000000007;
const long inv2 = 500000004;
int n;
vector<int> g[200000];
long sz[200000];

int rec(int from, int prev) {
  int sum = 1;
  for (int to : g[from]) {
    if (to == prev) continue;
    sum += rec(to, from);
  }
  sz[from] = sum;
  return sum;
}

long p[200001];
int main() {
  cin >> n;
  for (int i = 1; i < n; ++i) {
    int a, b;
    cin >> a >> b;
    --a; --b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  
  p[0] = 1;
  for (int i = 0; i < n; ++i) {
    p[i + 1] = p[i] * inv2 % mod;
  }
  
  rec(0, -1);
  long sum = 0;
  for (int i = 1; i < n; ++i) {
    long a = (1 + mod - p[sz[i]]) % mod * (1 + mod - p[n - sz[i]]) % mod;
    sum = (sum + a) % mod;
  }
  sum = (sum + mod - n * inv2 % mod) % mod;
  sum = (sum + mod - p[n] + 1) % mod;
  cout << sum << endl;
}
```