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>
#define ll long long
using namespace std;
const int MAXN=60;
int n,m,x,y,eg,u[MAXN],v[MAXN],hd[MAXN],ver[2*MAXN],nx[2*MAXN],f[MAXN];
bitset <51> q[21],t[1<<20];
ll ans;
ll qpow (ll a,ll b) {
	ll res=1;
	while (b) {
		if (b&1) {res*=a;}
		a*=a,b>>=1;
	}
	return res;
}
void add_edge (int x,int y) {
	ver[++eg]=y;
	nx[eg]=hd[x];
	hd[x]=eg;
	return;
}
void dfs (int x,int fa) {
	f[x]=fa;
	for (int i=hd[x];i;i=nx[i]) {
		if (ver[i]==fa) {continue;}
		dfs(ver[i],x);
	}
	return;
}
void mk (int x,int y,int p) {
	for (int i=x;i!=1;i=f[i]) {q[p].set(i,1);}
	int cur=y;
	for (;cur!=1&&!q[p][cur];cur=f[cur]) {q[p].set(cur,1);}
	for (;cur!=1;cur=f[cur]) {q[p].set(cur,0);}
	return;
}
int main () {
	eg=1;
	scanf("%d",&n);
	for (int i=1;i<=n-1;i++) {
		scanf("%d%d",&x,&y);
		add_edge(x,y),add_edge(y,x);
	}
	dfs(1,0);
	scanf("%d",&m);
	for (int i=1;i<=m;i++) {
		scanf("%d%d",&u[i],&v[i]);
		mk(u[i],v[i],i);
	}
	for (int i=0;i<(1<<m);i++) {
		int cnt=0,cur=0;
		for (int j=0;j<m;j++) {
			if (i&(1<<j)) {cnt++,cur=j+1;}
		}
		if (cnt) {t[i]=t[i^(1<<(cur-1))]|q[cur];}
		if (cnt&1) {
			ans-=qpow(2,n-1-t[i].count());
		} else {
			ans+=qpow(2,n-1-t[i].count());
		}
	}
	printf("%lld\n",ans);
	return 0;
}