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;

typedef long long ll;
#define fi first
#define se second
#define pb push_back

const int N = 55, M = 22;
set <int> st[M];
vector <pair <int, int> > v[N];
vector <int> path;
int a[N], b[N], n, m, rstrct, u1[M], u2[M];

void dfs(int x, int par) {
	if (x == u2[rstrct]) {
		for (int y : path) {
			st[rstrct].insert(y);
		}
		return;
	}
	for (auto y : v[x]) {
		if (y.fi != par) {
			path.pb(y.se);
			dfs(y.fi, x);
			path.pop_back();
		}
	}
}

bool vis[N];

ll rec(int x, int mask) {
	if (x == m) {
		memset(vis, 0, sizeof vis);
		int cnt = __builtin_popcount(mask), tot = n - 1;
		for (int i = 0; i < m; i++) {
			if ((1 << i) & mask) {
				for (int j : st[i]) {
					if (!vis[j]) {
						vis[j] = true;
						tot--;
					}
				}
			}
		}
		return (1LL << tot) * ((cnt & 1) ? -1 : 1);
	}
	return rec(x + 1, mask) + rec(x + 1, mask | (1 << x));
}

int main() {
    //ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    scanf("%d", &n);
    for (int i = 1; i < n; i++) {
		scanf("%d %d\n", &a[i], &b[i]);
		v[a[i]].pb({b[i], i});
		v[b[i]].pb({a[i], i});
	}
	scanf("%d", &m);
	for (int i = 0; i < m; i++) {
		scanf("%d %d", &u1[i], &u2[i]);
		rstrct = i;
		dfs(u1[i], u1[i]);
	}
	printf("%lld", rec(0, 0));
    return 0;
}