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

bitset<55> e[55][55], d[1 << 21];
int c[1 << 21];
int main() {
    int n;
    cin >> n;
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= n; j++)
            if(i != j)
                e[i][j] = ~e[i][j];
    for(int i = 1; i < n; i++) {
        int a, b;
        cin >> a >> b;
        e[a][b] = e[b][a] = 0;
        e[a][b][i] = e[b][a][i] = 1;
    }
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= n; j++)
            for(int k = 1; k <= n; k++) {
                auto x = e[j][i] | e[i][k];
                if(x.count() < e[j][k].count())
                    e[j][k] = x;
            }
    int m;
    cin >> m;
    for(int i = 0; i < m; i++) {
        int u, v;
        cin >> u >> v;
        d[1 << i] = e[u][v];
    }
    for(int i = 1; i < 1 << m; i++)
        d[i] = d[i & i - 1] | d[i & -i], c[i] = c[i & i - 1] + 1;
    long r = 0;
    for(int i = 0; i < 1 << m; i++)
        r += (c[i] & 1 ? -1 : 1) * (1l << n - 1 - d[i].count());
    cout << abs(r) << endl;
    return 0;
}
