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 MAXN 55

using ll = long long;

ll g[MAXN], path[MAXN][MAXN];
map<pair<int, int>, int> id;

void dfs(int src, int u, ll cur, ll p) {
	cur |= 1LL<<u;
	
	for(int b = 0; b < MAXN; b++) {
		if(g[u] >> b & 1) {
			if(cur >> b & 1) continue;
			int x = min(u, b), y = max(u, b);
			p ^= 1LL<<id[make_pair(x, y)];
			path[src][b] = p;
			dfs(src, b, cur, p);
			p ^= 1LL<<id[make_pair(x, y)];
		}
	}
}

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	
	int n;
	cin >> n;
	for(int i = 0; i < n - 1; i++) {
		int u, v;
		cin >> u >> v;
		u--, v--;
		g[u] |= 1LL<<v;
		g[v] |= 1LL<<u;
		if(u > v) swap(u, v);
		id[make_pair(u, v)] = i;
	}
	
	int m;
	cin >> m;
	vector<pair<int, int>> restrict(m);
	for(auto& p : restrict) {
		cin >> p.first >> p.second;
		p.first--, p.second--;
	}
	
	ll ans = 1LL<<(n-1);
	
	for(int i = 0; i < n; i++) {
		dfs(i, i, 0LL, 0LL);
	}
	
	for(ll s = 1; s < (1LL<<m); s++) {
		int add_sub = __builtin_popcountll(s) % 2 ? -1 : 1;
		
		ll edges = 0;
		for(int b = 0; b < m; b++) {
			if(s >> b & 1) {
				int u = restrict[b].first, v = restrict[b].second;
				edges |= path[u][v];
			}
		}
		
		ans += add_sub * (1LL<<(n - 1 - __builtin_popcountll(edges)));
	}
	
	cout << ans << endl;
}
