Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n,m,x[55];
vector<pair<int,int>> v[55];
long long dp[1<<20];
bool dfs1(int a,int b,int c,int d){
	if (a==c)
		return 1;
	for (auto& i : v[a])
		if (i.first!=b && dfs1(i.first,a,c,d)){
			x[i.second]^=1<<d;
			return 1;
		}
	return 0;
}
int main(){
	scanf("%d",&n);
	for (int i=1;i<n;i++){
		int ta,tb;
		scanf("%d%d",&ta,&tb);
		v[ta].emplace_back(tb,i);
		v[tb].emplace_back(ta,i);
	}
	scanf("%d",&m);
	for (int i=0;i<m;i++){
		int ta,tb;
		scanf("%d%d",&ta,&tb);
		dfs1(ta,0,tb,i);
	}
	dp[0]=1;
	for (int i=1;i<n;i++)
		for (int j=(1<<m)-1;j>=0;j--)
			dp[j|x[i]]+=dp[j];
	printf("%lld\n",dp[(1<<m)-1]);///////////////////////////////
}