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<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
typedef long long ll;
const int maxn = 105;
int n, m;
int head[maxn], nxt[maxn << 1], tail[maxn << 1], ecnt = 1;
int a[maxn], x[maxn], y[maxn];
bool col[maxn << 1];
ll ans;

void addedge(int u, int v) {
	nxt[++ecnt] = head[u];
	head[u] = ecnt;
	tail[ecnt] = v;
}

bool dfs1(int u, int p, int t) {
	if (u == t) return true;
	for (int e = head[u]; e; e = nxt[e]) {
		int v = tail[e];
		if (v != p)
			if (dfs1(v, u, t)) {
				col[e] = col[e ^ 1] = true;
				return true;
			}
	}
	return false;
}

void dfs(int pos, int cnt) {
	if (pos == m + 1) {
		if (!cnt) return ;
		memset(col, false, sizeof(col));
		for (int i = 1; i <= cnt; i++) dfs1(x[a[i]], 0, y[a[i]]);
		ll tmp = ecnt - 1; // * 2
		for (int i = 2; i <= ecnt; i++) tmp -= col[i];
		tmp >>= 1;
		tmp = (1LL << tmp);
		if (cnt & 1) ans -= tmp;
		else ans += tmp;
		return ;
	}
	dfs(pos + 1, cnt);
	a[cnt + 1] = pos;
	dfs(pos + 1, cnt + 1);
	a[cnt + 1] = 0;
}

int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n - 1; i++) {
		int u, v; scanf("%d%d", &u, &v);
		addedge(u, v); addedge(v, u);
	}
	scanf("%d", &m);
	for (int i = 1; i <= m; i++) scanf("%d%d", &x[i], &y[i]);
	ans = (1LL << (n - 1));
	dfs(1, 0);
	printf("%lld\n", ans);
	return 0;
}