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 sz(x) ((int)x.size())
#define all(x) (x).begin(), (x).end()
#define fi first
#define se second
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<vi> vvi;
int n,m;
int G[52][52];
int path[52],top;
bool chk[52];
vector<int> lst[52];
bool dfs(int p,int u,int to) {
	if(u==to) return true;
	for(int v=1;v<=n;v++) {
		if(!G[u][v]||v==p) continue;
		path[top++]=G[u][v];
		if(dfs(u,v,to)) return true;
		top--;
	}
	return false;
}
int main() {
	int i,j;
	scanf("%d",&n);
	for(i=1;i<=n-1;i++) {
		int a,b;
		scanf("%d%d",&a,&b);
		G[a][b]=G[b][a]=i;
	}
	scanf("%d",&m);
	for(i=0;i<m;i++) {
		int a,b;
		scanf("%d%d",&a,&b);
		top=0;
		dfs(a,a,b);
		for(j=0;j<top;j++) lst[i].push_back(path[j]);
	}
	ll res=0;
	for(i=0;i<1<<m;i++) {
		memset(chk,0,sizeof(chk));
		int c=0;
		for(j=0;j<m;j++) {
			if(i>>j&1) {
				c++;
				for(int k=0;k<sz(lst[j]);k++) {
					chk[lst[j][k]]=true;
				}
			}
		}
		int t=n-1;
		for(j=1;j<=n-1;j++) if(chk[j]) t--;
		if(c%2) res-=1ll<<t;
		else res+=1ll<<t;
	}
	printf("%lld",res);
	return 0;
}