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;
#define ll long long

const int N = 55;
int n, m, fa[N], dep[N];
ll ans, s[22];
vector<int> e[N];
void dfs(int u){
	for(int v:e[u]) if(v!=fa[u]) fa[v]=u, dep[v]=dep[u]+1, dfs(v);
}
int main() {
	scanf("%d", &n);
	for(int i=1, x, y; i<n; ++i)
		scanf("%d%d", &x, &y), e[x].push_back(y), e[y].push_back(x);
	dfs(1);
	scanf("%d", &m);
	for(int i=0, x, y; i<m; ++i){
		scanf("%d%d", &x, &y);
		if(dep[x]<dep[y]) swap(x, y);
		while(dep[x]>dep[y]) s[i]|=1ll<<x>>1, x=fa[x];
		while(x!=y) s[i]|=(1ll<<x>>1)|(1ll<<y>>1), x=fa[x], y=fa[y];
	}
	for(int i=0; i<1<<m; ++i){
		int cnt=0, cnt2=0;
		ll ss=0;
		for(int j=0; j<m; ++j) if(i>>j&1) ++cnt, ss|=s[j];
		for(int j=0; j<n; ++j) if(ss>>j&1) ++cnt2;
		ans+=(cnt&1?-1:1)*(1ll<<(n-1-cnt2));
	}
	printf("%lld\n", ans);
	return 0;
}