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 <functional>
#include <cstring>
#include <vector>
#include <iostream>
using namespace std;
int main() {
    int n; cin >> n;
    struct edge { int src, dst, C; } edges[n-1];
    for (int i = 0; i < n-1; i++) {
        int u, v; cin >> u >> v;
        edges[i] = {u-1, v-1, 0};
    }
    int m; cin >> m;
    int C[n];
    memset(C, 0, sizeof C);
    function <bool(int, int, int, int)> dfs = [&](int i, int j, int c, int p) {
        for (edge &e : edges) {
            int k = e.src == i ? e.dst : e.dst == i ? e.src : -1;
            if (k == -1 || k == p) continue;
            if (k == j || dfs(k, j, c, i)) {
                e.C |= (1<<c);
                return true;
            }
        }
        return false;
    };
    for (int i = 0; i < m; i++) {
        int u, v; cin >> u >> v;
        dfs(u-1, v-1, i, -1);
    }
    long long dp[1<<m];
    memset(dp, 0, sizeof dp);
    dp[0] = 1;
    function <void(int, int)> dfs2 = [&] (int i, int p) {
        for (edge &e : edges) {
            int k = e.src == i ? e.dst : e.dst == i ? e.src : -1;
            if (k == -1 || k == p) continue;
            for (int s = (1<<m)-1; s >= 0; s--) dp[s|e.C] += dp[s];
            dfs2(k, i);
        }
    };
    dfs2(0, -1);
    cout << dp[(1<<m)-1] << endl;
}
