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 rep(i,n) for(int i=0;i<(n);i++)

using namespace std;
using lint=long long;

inline int popcount(long long x){
	x-=(x>>1)&0x5555555555555555LL;
	x=(x&0x3333333333333333LL)+((x>>2)&0x3333333333333333LL);
	x=(x+(x>>4))&0x0f0f0f0f0f0f0f0fLL;
	x=(x*0x0101010101010101LL)>>56;
	return x;
}

template<class T> struct edge{
	int to;
	T wt;
	edge(int to,const T& wt):to(to),wt(wt){}
};
template<class T> using weighted_graph=vector<vector<edge<T>>>;

weighted_graph<int> T;

pair<bool,lint> dfs(int u,int p,int g){
	if(u==g){
		return {true,0};
	}

	for(auto e:T[u]) if(e.to!=p) {
		auto tmp=dfs(e.to,u,g);
		if(tmp.first){
			return {true,(1LL<<e.wt)|tmp.second};
		}
	}
	return {false,0};
}

int main(){
	int n; scanf("%d",&n);
	T.resize(n);
	rep(i,n-1){
		int u,v; scanf("%d%d",&u,&v); u--; v--;
		T[u].emplace_back(v,i);
		T[v].emplace_back(u,i);
	}

	int m; scanf("%d",&m);
	lint E[20];
	rep(i,m){
		int a,b; scanf("%d%d",&a,&b); a--; b--;
		E[i]=dfs(a,-1,b).second;
	}

	lint ans=0;
	rep(S,1<<m){
		lint white=0;
		rep(i,m) if(S>>i&1) white|=E[i];
		int e=popcount(white);
		ans+=(popcount(S)%2==0?1:-1)*(1LL<<(n-1-e));
	}
	printf("%lld\n",ans);

	return 0;
}
