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;
#define rep(i,n) for(int i=0; i<(n); i++)
using P = pair<int,int>;
using ll = long long;
int n, m, a, b;
ll ans, pw[55];
unsigned long long bt[55];
vector<P> G[55];
bool dfs(int u, int p, int t, int i){
	for(P nx: G[u]) if(nx.first != p){
		if(nx.first == t || dfs(nx.first,u,t,i)){
			bt[i] |= 1ll << nx.second;
			return true;
		}
	}
	return false;
}
int main(){
	scanf("%d", &n);
	rep(i,n-1){
		scanf("%d%d", &a, &b);
		G[a].push_back(P{b,i});
		G[b].push_back(P{a,i});
	}
	scanf("%d", &m);
	rep(i,m){
		scanf("%d%d", &a, &b);
		dfs(a,a,b,i);
	}
	pw[0] = 1;
	rep(i,n-1) pw[i+1] = pw[i] * 2;
	rep(i,1<<m){
		unsigned long long tmp = 0;
		rep(j,m) if(i>>j&1) tmp |= bt[j];
		ans += pw[n-1-__builtin_popcountll(tmp)] * (__builtin_popcount(i)%2 ? -1 : 1);
	}
	printf("%lld\n", ans);
}