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 i64 long long int

int n, m;
vector<pair<int, int> > e[55];

vector<int> t, p[22];

i64 pw[55];
i64 ans;

bool dfs_path(int cur, int par, int ed) {
	if(cur==ed) return true;
	
	for(int i=0;i<(int)e[cur].size();i++) {
		int nxt = e[cur][i].first;
		if(nxt==par) continue;
		
		t.push_back(e[cur][i].second);
		if(dfs_path(nxt, cur, ed)) return true;
		
		t.pop_back();
	}
	return false;
}

vector<int> get_path(int st, int ed) {
	t.clear();
	dfs_path(st, -1, ed);
	return t;
}

void dfs_calc(int id, set<int> s, bool flag) {
	if(id==m) {
		if(flag) ans -= pw[n-1-s.size()];
		else ans += pw[n-1-s.size()];
		
		return;
	}
	
	dfs_calc(id+1, s, flag);
	for(int i=0;i<(int)p[id].size();i++)
		s.insert(p[id][i]);
	dfs_calc(id+1, s, 1-flag);
}

int main() {
	scanf("%d", &n);
	for(int i=1;i<n;i++) {
		int x,y;
		scanf("%d%d", &x, &y);
		e[x].push_back(make_pair(y, i-1));
		e[y].push_back(make_pair(x, i-1));
	}
	scanf("%d", &m);
	for(int i=0;i<m;i++) {
		int x,y;
		scanf("%d%d", &x, &y);
		p[i] = get_path(x, y);
	}
	
	pw[0] = 1LL;
	for(int i=1;i<=n;i++)
		pw[i] = pw[i-1]*2;
	
	ans = 0LL;
	set<int> s; s.clear();
	dfs_calc(0, s, 0);
	cout<<ans<<endl;
	
	return 0;  
}