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

// Problem : F - Tree and Constraints
// Contest : AtCoder Beginner Contest 152
// URL : https://atcoder.jp/contests/abc152/tasks/abc152_f
// Memory Limit : 1024.000000 MB 
// Time Limit : 4000.000000 milisec 
// Powered by CP Editor (https://github.com/coder3101/cp-editor)

#include <bits/stdc++.h>

using namespace std;

int N, M;
int u[55], v[55];
vector<int> graph[55];
int msk[55];
long long dp[55][1<<20];
int lft[55], rht[55], t;
int dep[55], par[55];

void dfs(int n){
	lft[n] = ++t;
	for(auto e : graph[n]){
		if(e != par[n]){
			par[e] = n;
			dep[e] += dep[n] + 1;
			dfs(e);
		}
	}
	rht[n] = ++t;
}

int main(){
	cin.sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	cin >> N;
	for(int i = 1; i<N; i++){
		cin >> u[i] >> v[i];
		graph[u[i]].push_back(v[i]);
		graph[v[i]].push_back(u[i]);
	}
	dfs(1);
	cin >> M;
	for(int j = 0; j<M; j++){
		int a, b;
		cin >> a >> b;
		for(int i = 1; i<N; i++){
			int c = 0;
			if(par[u[i]] == v[i]){
				swap(u[i], v[i]);
			}
			c += (lft[v[i]] <= lft[a] && rht[a] <= rht[v[i]]);
			c += (lft[v[i]] <= lft[b] && rht[b] <= rht[v[i]]);
			if(c == 1){
				msk[i] |= (1<<j);
			}
		}
	}
	dp[0][0] = 1;
	for(int i = 0; i<N-1; i++){
		for(int m = 0; m<1<<M; m++){
			dp[i+1][m] += dp[i][m];
			dp[i+1][m|msk[i+1]] += dp[i][m];
		}
	}
	cout << dp[N-1][(1<<M)-1] << "\n";
}