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;
long long a,b,c,d,e,m,i,j,lc[59],dep[59],msh[59],pas,k[59],zx,xc,xr[59];
pair <long long, long long> p[59];
vector <long long> v[59];
void dfsst(long long q, long long w){
	msh[q]=w;
	dep[q]=dep[w]+1;
	for(vector <long long>::iterator it=v[q].begin(); it!=v[q].end(); it++){
		if((*it)==w) continue;
		dfsst((*it),q);
	}
}
void dfs(long long q, long long w){
	for(vector <long long>::iterator it=v[q].begin(); it!=v[q].end(); it++){
		if((*it)==w) continue;
		dfs((*it),q);
		k[q]+=k[(*it)];
	}
}
int main(){
	ios_base::sync_with_stdio(false),cin.tie(0),cout.tie(0);
	cin>>a;
	for(b=1; b<a; b++){
		cin>>c>>d;
		v[c].push_back(d);
		v[d].push_back(c);
	}
	dfsst(1,0);
	xr[0]=1;
	for(i=1; i<=a; i++) xr[i]=xr[i-1]*2;
	cin>>m;
	for(i=0; i<m; i++){
		cin>>p[i].first>>p[i].second;
		c=p[i].first;d=p[i].second;
		while(c!=d){
			if(dep[c]>=dep[d]){
				c=msh[c];
			}else{
				d=msh[d];
			}
		}
		lc[i]=c;
	}
	for(i=1; i<(1<<m); i++){
		for(j=1; j<=a; j++) k[j]=0;
		for(j=0; j<m; j++){
			if((i&(1<<j))==0) continue;
			k[p[j].first]++;
			k[p[j].second]++;
			k[lc[j]]-=2;
		}
		dfs(1,0);
		zx=0;
		for(j=2; j<=a; j++) if(k[j]==0) zx++;
		if(__builtin_popcount(i)%2==1) pas+=xr[zx]; else pas-=xr[zx];
	}
	cout<<xr[a-1]-pas;
	return 0;
}