We have a tree with N vertices numbered 1 to N. The i-th edge in this tree connects Vertex a_i and Vertex b_i.
Consider painting each of these edges white or black. There are 2^{N-1} such ways to paint the edges. Among them, how many satisfy all of the following M restrictions?

* The i-th (1 \leq i \leq M) restriction is represented by two integers u_i and v_i, which mean that the path connecting Vertex u_i and Vertex v_i must contain at least one edge painted black.

Constraints

* 2 \leq N \leq 50
* 1 \leq a_i,b_i \leq N
* The graph given in input is a tree.
* 1 \leq M \leq \min(20,\frac{N(N-1)}{2})
* 1 \leq u_i < v_i \leq N
* If i \not= j, either u_i \not=u_j or v_i\not=v_j
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N
a_1 b_1
:
a_{N-1} b_{N-1}
M
u_1 v_1
:
u_M v_M


Output

Print the number of ways to paint the edges that satisfy all of the M conditions.

Examples

Input

3
1 2
2 3
1
1 3


Output

3


Input

2
1 2
1
1 2


Output

1


Input

5
1 2
3 2
3 4
5 3
3
1 3
2 4
2 5


Output

9


Input

8
1 2
2 3
4 3
2 5
6 3
6 7
8 6
5
2 7
3 5
1 6
2 8
7 8


Output

62
#include <iostream>
#include <vector>
#include <functional>
using namespace std;

int main()
{
    int N;
    cin>>N;
    vector<vector<int>> E(N), EI(N);
    for (int i=0; i<N-1; i++)
    {
        int a, b;
        cin>>a>>b;
        E[a-1].push_back(b-1);
        EI[a-1].push_back(i);
        E[b-1].push_back(a-1);
        EI[b-1].push_back(i);
    }

    function<long long(int, int, int, long long)> f = [&](int c, int p, int g, long long m)
    {
        if (c==g)
            return m;

        for (int i=0; i<(int)E[c].size(); i++)
            if (E[c][i]!=p)
            {
                long long r = f(E[c][i], c, g, m|1LL<<EI[c][i]);
                if (r!=-1)
                    return r;
            }
        return -1LL;
    };

    int M;
    cin>>M;
    vector<long long> C(M);
    for (long long &c: C)
    {
        int u, v;
        cin>>u>>v;
        c = f(u-1, -1, v-1, 0);
    }

    vector<long long> T(1<<M);
    T[0] = 1;
    for (int i=0; i<N-1; i++)
    {
        int x = 0;
        for (int j=0; j<M; j++)
            if (C[j]>>i&1)
                x |= 1<<j;

        vector<long long> P = T;
        for (int i=0; i<1<<M; i++)
            T[i|x] += P[i];
    }
    cout<<T[(1<<M)-1]<<endl;
}
