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;

int pop_count(int bits){
    bits = (bits & 0x55555555) + (bits >> 1 & 0x55555555);
    bits = (bits & 0x33333333) + (bits >> 2 & 0x33333333);
    bits = (bits & 0x0f0f0f0f) + (bits >> 4 & 0x0f0f0f0f);
    bits = (bits & 0x00ff00ff) + (bits >> 8 & 0x00ff00ff);
    return (bits & 0x0000ffff) + (bits >>16 & 0x0000ffff);
}

int N, M;
vector<pair<int, int>> edges[50];

int mask[50];
bool dfs_mask(int i, int p, int target, int ei, int b){
    bool ok = (i==target);
    for(auto& e : edges[i]) if(e.first != p) ok |= dfs_mask(e.first, i, target, e.second, b);
    if(ok && ei >= 0) mask[ei] |= 1<<b;
    return ok;
}

int main(){
    cin >> N;
    for(int i=0; i<N-1; i++){
        int a, b;
        cin >> a >> b;
        edges[a-1].emplace_back(b-1, i);
        edges[b-1].emplace_back(a-1, i);
    }
    cin >> M;
    for(int i=0; i<M; i++){
        int a, b;
        cin >> a >> b;
        dfs_mask(a-1, -1, b-1, -1, i);
    }
    int64_t ans = 0;
    for(int b=0; b<(1<<M); b++){
        int64_t res = 1;
        for(int i=0; i<N-1; i++) if(!(b & mask[i])) res *= 2;
        if(pop_count(b)%2) res *= -1;
        ans += res;
    }
    cout << ans << endl;
    return 0;
}
