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;

vector<vector<array<int, 2>>> E;

int64_t dfs(int i, int goal, int from){
    if (i == goal)
        return 0;
    for (const auto &e: E[i]){
        if (e[0] == from)
            continue;
        auto res = dfs(e[0], goal, i);
        if (res >= 0)
            return res | (1L << e[1]);
    }
    return -1;
}

int main(){
    int n;
    cin >> n;
    E.resize(n);
    for (int i = 0; i < n - 1; i++){
        int a, b;
        cin >> a >> b;
        a--; b--;
        E[a].push_back({b, i});
        E[b].push_back({a, i});
    }
    int m;
    cin >> m;
    vector<int> U(m), V(m);
    for (int i = 0; i < m; i++){
        cin >> U[i] >> V[i];
        U[i]--; V[i]--;
    }

    vector<int64_t> masks(m);
    for (int i = 0; i < m; i++){
        masks[i] = dfs(U[i], V[i], -1);
    }

    int64_t ans = (1L << (n - 1));
    for (int bits = 1; bits < (1 << m); bits++){
        int64_t mask = 0;
        for (int i = 0; i < m; i++){
            if (bits >> i & 1)
                mask |= masks[i];
        }
        int c = __builtin_popcount(bits) & 1 ? -1 : 1;
        ans += c * (1L << (n - 1 - __builtin_popcountl(mask)));
    }
    
    cout << ans << endl;
}