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 ll = long long;

const int N = 55, M = 21;

int n;
vector<pair<int, int>> g[N];
int m;
int pt[N];

bool prec(int st, int en, int id, int par = -1) {
	if(st == en) return true;
	for(auto p : g[st]) if(p.first != par) {
		if(prec(p.first, en, id, st)) {
			pt[p.second] |= 1 << id;
			return true;
		}
	}
	return false;
}

ll dp[N][1 << M];

ll f(int i, int mask) {
	if(i == n - 1) return mask == (1 << m) - 1;
	ll &ans = dp[i][mask];
	if(~ans) return ans;
	return ans = f(i + 1, mask) + f(i + 1, mask | pt[i]);
}

int main() {
	memset(dp, -1, sizeof dp);
	scanf("%d", &n);

	for(int i = 0; i < n - 1; i++) {
		int u, v;
		scanf("%d %d", &u, &v);
		g[u].emplace_back(v, i);
		g[v].emplace_back(u, i);
	}

	scanf("%d", &m);
	for(int i = 0; i < m; i++) {
		int u, v;
		scanf("%d %d", &u, &v);
		assert(prec(u, v, i));
	}

	printf("%lld\n", f(0, 0));
}
