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 <algorithm>
#include <vector>
#include <queue>
#include <string>

using ll = long long;

using namespace std;

using P = pair<int, int>;
#define fi first
#define se second

vector<P> G[51]; // 行先, id
bool visited[51] = {0};
int edges[51] = {0};

bool dfs(int i, int to, int id) {
    visited[i] = true;
    if (i == to) 
        return true;
    for (auto &g:G[i]) {
        if (visited[g.fi]) continue;
        int d = dfs(g.fi, to, id);
        if (d) {
            edges[g.se] |= (1 << id); // 辺g.seによって条件idが成立する
            return true;
        }
    }
    return false;
}

ll dp[51][(1 << 20) + 1];

int main() {
    int n, m;
    cin >> n;
    for (int i=0, ai, bi; i<n-1; i++) {
        cin >> ai >> bi;
        ai--; bi--;
        G[ai].push_back({bi, i});
        G[bi].push_back({ai, i});
    }
    cin >> m;
    for (int i=0, ui, vi; i<m; i++) {
        cin >> ui >> vi;
        ui--; vi--;
        fill(visited, visited+n, false);
        dfs(ui, vi, i);
    }
    fill(dp[0], dp[n], 0);
    dp[0][0] = 1;
    for (int i=0; i<n-1; i++) {
        for (int j=0; j<(1<<m); j++)
            dp[i+1][j] = dp[i][j];
        for (int j=0; j<(1 << m); j++) {
            dp[i+1][j | edges[i]] += dp[i][j];
        }
    }
    cout << dp[n-1][(1 << m) - 1] << '\n';
    return 0;
}