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;

typedef long long ll;

const ll mod = 1e9 + 7;

ll n;
vector<int> g[200005];
ll ans;

ll p(ll a, ll x)
{
    ll res = 1;

    while (x > 0)
    {

        if (x & 1)
        {
            res = a * res % mod;
        }

        a = a * a % mod;
        x >>= 1;
    }
    return res;
}

ll dfs(int v, int prev = -1)
{
    //cout << "iji" << endl;
    ll s = 1;
    vector<ll> t;

    for (int i = 0; i < g[v].size(); i++)
    {
        int u = g[v][i];
        if (u == prev)
        {
            continue;
        }

        ll tmp = dfs(u, v);
        t.push_back(tmp);
        s += tmp;
    }

    t.push_back(n - s);

    if (t.size() == 1)
        return s;

    ll res = (p(2, n - 1) - 1) % mod;

    for (int i = 0; i < t.size(); i++)
    {
        res = (res - p(2, t[i]) + 1 + mod) % mod;
    }
    ans = (ans + res) % mod;
    //cout << "v = " << v << "  ans = " << ans << endl;
    return s;
}

int main()
{
    cin >> n;

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

    dfs(0);

    cout << ans * p(p(2, n), mod - 2) % mod << endl;
}
