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>
#define long long long int
using namespace std;

// @author: pashka

vector<vector<pair<int, int>>> g;
vector<bool> mark;

bool dfs(int x, int y, int p) {
    if (x == y) return true;
    for (auto e : g[x]) {
        if (e.first == p) continue;
        if (dfs(e.first, y, x)) {
            mark[e.second] = true;
            return true;
        }
    }
    return false;
}

int main() {
    ios::sync_with_stdio(false);

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

    int m;
    cin >> m;
    vector<int> cover(n - 1);
    for (int i = 0; i < m; i++) {
        int x, y;
        cin >> x >> y;
        x--; y--;
        mark.assign(n - 1, false);
        dfs(x, y, -1);
        for (int j = 0; j < n - 1; j++) {
            if (mark[j]) {
                cover[j] |= (1 << i);
            }
        }
    }

    vector<long> d(1 << m);
    d[0] = 1;
    for (int i = 0; i < n - 1; i++) {
        for (int x = (1 << m) - 1; x >= 0; x--) {
            long q = d[x];
            d[x | cover[i]] += q;
        }
    }
    cout << d[(1 << m) - 1];

    return 0;
}