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;
using Int = int_fast64_t;

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);
	Int n; cin >> n;
	vector<vector<Int>> g(n);
	for(Int i=0; i<n-1; ++i){
		Int a, b; cin >> a >> b; --a; --b;
		g[a].push_back(b);
		g[b].push_back(a);
	}
	vector<Int> p(n, -1), d(n, -1);
	p[0] = -1;
	d[0] = 0;
	queue<Int> q;
	q.push(0);
	while(!q.empty()){
		Int v = q.front(); q.pop();
		for(auto t:g[v])
			if(d[t] == -1){
				p[t] = v;
				d[t] = d[v] + 1;
				q.push(t);
			}
	}
	Int m; cin >> m;
	vector<Int> u(m), v(m);
	for(Int i=0; i<m; ++i){
		cin >> u[i] >> v[i];
		--u[i]; --v[i];
	}
	vector<Int> l(m);
	for(Int i=0; i<m; ++i){
		Int x = u[i], y = v[i];
		if(d[x] < d[y])
			swap(x, y);
		while(d[x] > d[y])
			x = p[x];
		while(x != y){
			x = p[x];
			y = p[y];
		}
		l[i] = x;
	}
	Int ans = 0;
	vector<Int> w;
	for(Int b=0; b<(1ll<<m); ++b){
		Int cnt = 0;
		w.assign(n, 0);
		for(Int i=0; i<m; ++i)
			if((b >> i) & 1){
				++cnt;
				Int x = u[i], y = v[i];
				while(x != l[i]){
					w[x] = 1;
					x = p[x];
				}
				while(y != l[i]){
					w[y] = 1;
					y = p[y];
				}
			}
		Int nw = 0;
		for(Int i=0; i<n; ++i)
			nw += w[i];
		ans += (1-2*(cnt%2)) * (1ll << (n-1-nw));
	}
	cout << ans << "\n";
}