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;

const int maxn=55,maxm=22;

long long dp[maxn][1<<maxm];
int n,m,msk[maxn];
vector <pair<int,int> > g[maxn];

bool dfs(int id,int x,int fa,int ed){
	if(x==ed) return true;
	for(int i=0;i<g[x].size();i++){
		int to=g[x][i].first,ind=g[x][i].second;
		if(to==fa) continue;
		if(dfs(id,to,x,ed)){
			msk[ind]|=1<<id;
			return true;
		}
	}
	return false;
}

int main(){
	scanf("%d",&n);
	for(int i=0;i<n-1;i++){
		int a,b;
		scanf("%d%d",&a,&b);
		a--; b--;
		g[a].push_back(make_pair(b,i));
		g[b].push_back(make_pair(a,i));
	}
	scanf("%d",&m);
	for(int i=0;i<m;i++){
		int a,b;
		scanf("%d%d",&a,&b);
		a--; b--;
		dfs(i,a,-1,b);
	}
	dp[0][0]=1;
	for(int i=0;i<n-1;i++){
		for(int mask=0;mask<(1<<m);mask++){
			dp[i+1][mask]+=dp[i][mask];
			dp[i+1][mask|msk[i]]+=dp[i][mask];
		}
	}
	printf("%lld\n",dp[n-1][(1<<m)-1]);
	return 0;
}