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;
vector<int>g[59];
int n,logn,depth[59],parent[9][59],m;
long long state[29],ans;
void dfs(int u,int p,int d) {
	parent[0][u]=p;
	depth[u]=d;
	for(int i=0;i<g[u].size();i++)
		if(g[u][i]!=p)
			dfs(g[u][i],u,d+1);
}
int lca(int u,int v) {
	if(depth[u]>depth[v]) swap(u,v);
	for(int i=0;i<=logn;i++)
		if((depth[v]-depth[u])>>i&1)
			v=parent[i][v];
	if(u==v) return u;
	for(int i=logn;i>=0;i--)
		if(parent[i][u]!=parent[i][v]) {
			u=parent[i][u];
			v=parent[i][v];
		}
	return parent[0][u];
}
int main() {
	scanf("%d",&n);
	logn=log2(n);
	ans=1ll<<(n-1);
	for(int i=0;i<n-1;i++) {
		int a,b;
		scanf("%d%d",&a,&b);
		g[a].push_back(b);
		g[b].push_back(a);
	}
	dfs(1,0,0);
	for(int i=1;i<=logn;i++)
		for(int j=1;j<=n;j++)
			parent[i][j]=parent[i-1][parent[i-1][j]];
	scanf("%d",&m);
	for(int i=0;i<m;i++) {
		int u,v;
		scanf("%d%d",&u,&v);
		int p=lca(u,v);
		for(;u!=p;u=parent[0][u]) state[i]|=1ll<<u;
		for(;v!=p;v=parent[0][v]) state[i]|=1ll<<v;
	}
	for(int mask=1;mask<(1<<m);mask++) {
		long long path=0;
		for(int i=0;i<m;i++)
			if((1<<i)&mask)
				path|=state[i];
		ans+=(1ll<<(n-1-__builtin_popcountll(path)))*(__builtin_popcount(mask)%2?-1ll:1ll);
	}
	printf("%lld\n",ans);
	return 0;
}