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 <iostream>
#include <string>
#include <algorithm>
#include <vector>
#include <map>
#include <cmath>
#include <queue>
#include <deque>
#include <set>
#include <iomanip>
#include <utility>
#define ll long long
#define ld long double
using namespace std;

vector<vector<int>> G;
int par[60], depth[60], num[60], cnt=1;
ll path[25];
void dfs(int key){
	for(int i=0; i<G[key].size(); ++i){
		if(par[key]!=G[key][i]){
			par[G[key][i]]=key;
			depth[G[key][i]]=depth[key]+1;
			num[G[key][i]]=cnt;
			++cnt;
			dfs(G[key][i]);
		}
	}
}

int main() {
	int N;
	cin >> N;
	G.resize(N+1);
	for(int i=1; i<N; ++i){
		int a, b;
		cin >> a >> b;
		G[a].emplace_back(b);
		G[b].emplace_back(a);
	}
	dfs(1);
	int M;
	cin >> M;
	for(int i=0; i<M; ++i){
		int u, v;
		cin >> u >> v;
		int now=u;
		while(now>1){
			path[i] ^= (ll) 1<<(num[now]);
			now=par[now];
		}
		now=v;
		while(now>1){
			path[i] ^= (ll) 1<<(num[now]);
			now=par[now];
		}
	}
	ll ans=(ll) 1<<(N-1);
	for(int i=1; i<(1<<M); ++i){
		ll sum=0;
		int ccnt=0;
		for(int d=0; d<M; ++d){
			if(i&(1<<d)){
				sum |= path[d];
				++ccnt;
			}
		}
		int dcnt=N-1;
		for(int d=1; d<N; ++d){
			if(sum&((ll) 1<<d)){
				--dcnt;
			}
		}
		if(ccnt%2==1) ans -= (ll) 1<<dcnt;
		else ans += (ll) 1<<dcnt;
	}
	cout << ans << endl;
	return 0;
}