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 <iostream>
#include <vector>

using namespace std;
vector<pair<int,int>> G[100];
vector<int> v,gr[50];
int ed[1100];
long long beki[60];
void dfs(int s, int p,int ind,int to){
	if(s==to){
		gr[ind] = v;
	}
	for(auto e:G[s]){
		if(e.first==p) continue;
		v.push_back(e.second);
		dfs(e.first,s,ind,to);
		v.pop_back();
	}
}

int main(){
	int i,j,k,n,m;
	cin >> n;
	for(i=0;i<n - 1;i++){
		int a,b; cin >> a >> b; a--; b--;
		G[a].push_back({b,i}); G[b].push_back({a,i});
	}
	cin >> m;
	for(i=0;i<m;i++){
		int a,b; cin >> a >> b; a--; b--;
		dfs(a,-1,i,b);
	}
	long long x = 1;
	for(i=0;i<n;i++){
		beki[i] = x;
		x *= 2;
	}
	long long ans = 0;
	for(i=0;i<(1<<m);i++){
		for(j=0;j<n - 1;j++){
			ed[j] = 0;
		}
		int cnt = 0,cnt1 = 0;
		for(j=0;j<m;j++){
			if((1<<j) & i){
				cnt1++;
				for(int k:gr[j]){
					ed[k] = 1;
				}
			}
		}
		for(j=0;j<n - 1;j++){
			cnt += ed[j];
		}
		if(cnt1&1) ans -= beki[n - 1 - cnt];
		else ans += beki[n - 1 - cnt];
	}
	cout << ans << endl;
}