Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int maxn = 105;
int n, m;
long long ans;
vector<int> G[maxn];
int fa[maxn], dep[maxn];
long long d[maxn];
map<pair<int, int>, int> mp;
void dfs(int u, int f) {
	fa[u] = f;
	for(auto v : G[u]) {
		if(v != f) {
			dep[v] = dep[u] + 1;
			dfs(v, u);
		}
	}
}
int main() {
	scanf("%d", &n);
	for(int i = 1; i < n; ++i) {
		int u, v; scanf("%d%d", &u, &v);
		G[u].push_back(v);
		G[v].push_back(u);
		mp[minmax(u, v)] = i - 1;
	}
	dfs(1, 0);
	scanf("%d", &m);
	for(int i = 0; i < m; ++i) {
		int u, v;
		scanf("%d%d", &u, &v);
		while(u != v) {
			if(dep[u] < dep[v]) {
				swap(u, v);
			}
			d[i] |= 1LL << mp[minmax(u, fa[u])];
			u = fa[u];
		}
	}
	for(int S = 0; S < 1 << m; ++S) {
		long long e = 0;
		for(int i = 0; i < m; ++i) {
			if(S >> i & 1) {
				e |= d[i];
			}
		}
		int c = 0;
		for(int i = 0; i < n - 1; ++i) {
			if(e & (1LL << i)) {
				++c;
			}
		}
		if(__builtin_popcount(S) & 1) {
			ans -= 1LL << (n - c - 1);
		} else {
			ans += 1LL << (n - c - 1);
		}
	}
	printf("%lld\n", ans);
	return 0;
}