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;
#ifdef DEBUG
#define FPEN freopen("in.txt", "r", stdin);
#define debug(...) \
  { cerr << "# " << __VA_ARGS__ << "\n"; }
#else
#define debug(...)
#define FPEN
#endif
const int maxn = 200009;
typedef long long i64;
typedef pair<i64, i64> pi64;
typedef pair<int, int> pi;
const i64 MOD = 1e9 + 7;
class {
 public:
  vector<int> V[maxn];
  i64 pw(i64 a, i64 b) {
    i64 c = 1;
    while (b) {
      if (b & 1)
        c = c * a % MOD;
      a = a * a % MOD;
      b >>= 1;
    }
    return c;
  }
  i64 P;
  i64 pw2[maxn];
  int n;
  void solve() {
    cin >> n;
    pw2[0] = 1;
    for (int i = 1; i < maxn; ++i) pw2[i] = pw2[i - 1] * 2 % MOD;
    P = 0;
    for (int i = 1; i < n; ++i) {
      int u, v;
      cin >> u >> v;
      V[v].push_back(u);
      V[u].push_back(v);
    }
    dfs(1, 1);
    P = P * pw(pw2[n], MOD - 2) % MOD;
    cout << (P % MOD + MOD) % MOD;
  }
  i64 sz[maxn];
  void dfs(int u, int f) {
    sz[u] = 1;
    for (int v : V[u]) {
      if (f != v) {
        dfs(v, u);
        sz[u] += sz[v];
      }
    }
    if (V[u].size() == 1)
      return;
    i64 szfa = n - sz[u];
    i64 res = pw2[n - 1] - 1;
    for (int v : V[u]) {
      if (v == f) {
        res = res - (pw2[szfa] - 1);
      } else {
        res = res - (pw2[sz[v]] - 1);
      }
      res = res % MOD;
    }
    P = (P + res) % MOD;
  }
} NSPACE;
int main() {
  FPEN;
  ios_base::sync_with_stdio(false);
  cout.tie(0);
  cin.tie(0);
  NSPACE.solve();
}