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;
typedef long long ll;
ll read(){
	ll a=0,b=getchar(),c=1;
	while(!isdigit(b))c=b=='-'?-1:1,b=getchar();
	while(isdigit(b))a=a*10+b-'0',b=getchar();
	return a*c;
}
vector<pair<ll,ll> > v[60];
ll a[60],M,dp[60][(1<<20)+5],n;
set<ll> st;
void dfs(ll x,ll p,ll y,ll j,ll idx){
	if(x==y){
		for(set<ll>::iterator i=st.begin();i!=st.end();i++)
			a[*i]|=(1<<j);
		return ;
	}
	for(int i=0;i<v[x].size();i++){
		pair<ll,ll> it=v[x][i];
		if(it.first!=p){
			st.insert(it.second);
			dfs(it.first,x,y,j,it.second);
		}
	}
	st.erase(idx);
}
ll gao(int i,ll y){
	if(i==n)return (y+1==(1<<M));
	if(dp[i][y]!=-1)return dp[i][y];
	ll res=0;
	res=gao(i+1,y|a[i])+gao(i + 1,y);
	return dp[i][y]=res;
}
int main(){
	memset(dp,-1,sizeof(dp));
	n=read();
	for(ll i=1;i<n;i++){
		ll x=read(),y=read();
		v[x].push_back({y,i});
		v[y].push_back({x,i});
	}
	M=read();
	for(ll i=0;i<M;i++){
		ll x=read(),y=read();
		dfs(x,-1,y,i,-1);
		st.clear();
	}
	cout << gao(1,0);
	return 0;
}