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;

vector<int> edge[55];
map<int, map<int, int>> idx;

bool dfs(int from, int par, int goal, bitset<55> &path) {
	if(from == goal) return true;
	bool ret = false;
	for(int to : edge[from]) {
		if(to == par) continue;
		bool tmp = dfs(to, from, goal, path);
		if(tmp) {
			ret = true;
			path.set(idx[from][to], true);
		}
	}
	return ret;
}

int main() {
	int n; cin >> n;
	for(int i = 0; i < n - 1; ++i) {
		int a, b; cin >> a >> b;
		--a, --b;
		edge[a].push_back(b);
		edge[b].push_back(a);
		idx[a][b] = idx[b][a] = i;
	}
	int m; cin >> m;
	bitset<55> path[m];
	for(int i = 0; i < m; ++i) {
		int u, v; cin >> u >> v;
		--u, --v;
		dfs(u, -1, v, path[i]);
	}
	int64_t ans = 0;
	for(int mask = 1; mask < (1 << m); ++mask) {
		bitset<55> OR;
		for(int i = 0; i < m; ++i) {
			if(mask >> i & 1) {
				OR |= path[i];
			}
		}
		int cnt = __builtin_popcount(mask);
		if(cnt % 2 == 0) {
			ans -= (1ll << (int64_t)(n - 1 - OR.count()));
		} else {
			ans += (1ll << (int64_t)(n - 1 - OR.count()));
		}
	}
	cout << (1ll << (n - 1)) - ans << '\n';
	return 0;
}