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 st first
#define nd second
using ll = long long;
using namespace std;
using pii = pair<int,int>;
map<pii,int> mp;
vector<int> gr[55];
ll bm[25];
ll dfs(int tar,int id,int fr){
    if(id == tar){
        return (1LL<<mp[{id,fr}]);
    }
    ll sum = 0LL;
    for(auto it: gr[id]){
        if(it != fr) sum += dfs(tar,it,id);
        if(sum > 0LL) break;
    }
    if(sum == 0LL) return sum;
    if(fr != 0) return sum + (1LL<<mp[{id,fr}]);
    return sum;
}
int main(){
    int n;
    scanf("%d",&n);
    int x,y;
    for(int i = 0;i<n-1;i++){
        scanf("%d%d",&x,&y);
        mp[{x,y}] = i;
        mp[{y,x}] = i;
        gr[x].push_back(y);
        gr[y].push_back(x);
    }
    int m;
    scanf("%d",&m);
    for(int i =0;i<m;i++){
        scanf("%d%d",&x,&y);
        bm[i] = dfs(y,x,0);
    }
    ll ans = 0LL;
    int ct = 1LL;
    for(int i = 0 ;i < (1<<m);i++){
        ll b = 0LL;
        ct = 1;
        for(int j = 0; j < m; j++){
            if(i & (1<<j)) b |= bm[j],ct=-ct;
        }
        int ma = n -1;
        for(int j = 0; j < n-1;j++){
            if(b & (1LL<<j)) ma--;
        }
        ans += ct * (1LL<<ma);
    }
    printf("%lld",ans);

}
