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;
typedef long long ll;
int n,m;
int a[55], b[55];
int x[25], y[25];
int ok[55];
int col[55];
vector<int> adjlist[55];
void dfs(int u){
    for (auto v : adjlist[u]){
        if (col[v] == -1){
            col[v] = col[u];
            dfs(v);
        }
    }
}
ll memo[55][(1<<20)];
ll dp(int x, int mask){
    if (x == n-1) return mask == ((1<<m)-1);
    if (memo[x][mask] != -1) return memo[x][mask];
    memo[x][mask] = dp(x+1,mask)+dp(x+1,mask|ok[x]);
    //printf("%d %d %d\n",x,mask,memo[x][mask]);
    return memo[x][mask];
}
int main(){
    scanf("%d",&n);
    for (int i = 0; i < n-1; i++){
        scanf("%d%d",&a[i],&b[i]);
        adjlist[a[i]].push_back(b[i]);
        adjlist[b[i]].push_back(a[i]);
    }
    scanf("%d",&m);
    for (int j = 0; j < m; j++){
        scanf("%d%d",&x[j],&y[j]);
    }
    for (int i = 0; i < n-1; i++){
        memset(col,-1,sizeof(col));
        col[a[i]] = 0;
        col[b[i]] = 1;
        dfs(a[i]);
        dfs(b[i]);
        for (int j = 0; j < m; j++){
            //printf("%d %d\n",col[x[j]],col[y[j]]);
            if (col[x[j]] != col[y[j]]) ok[i] |= (1<<j);
        }
    }
    memset(memo,-1,sizeof(memo));
    printf("%lld",dp(0,0));




}
