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>
using namespace std;
template<class t> inline t read(t &x){
    x=0;char c=getchar();bool f=0;
    while(!isdigit(c)) f|=c=='-',c=getchar();
    while(isdigit(c)) x=(x<<1)+(x<<3)+(c^48),c=getchar();
    if(f) x=-x;return  x;
}
template<class t> inline void write(t x){
    if(x<0){putchar('-'),write(-x);}
    else{if(x>9)write(x/10);putchar('0'+x%10);}
}

#define int long long

const int N=55,M=25;
int en,f[N],d[N],h[N],s[M],n,m,ans;

struct edge{
	int n,v;
}e[N<<1];

void add(int x,int y){
	e[++en]=(edge){h[x],y};
	h[x]=en;
}

void dfs(int x){
	d[x]=d[f[x]]+1;
	for(int i=h[x];i;i=e[i].n){
		int y=e[i].v;
		if(y==f[x]) continue;
		f[y]=x;
		dfs(y);
	}
}

signed main(){
	read(n);
	for(int i=1,x,y;i<n;i++){
		read(x);read(y);
		add(x,y);add(y,x);
	}
	dfs(1);
	read(m);
	for(int i=0,x,y;i<m;i++){
		read(x);read(y);
		if(d[x]<d[y]) swap(x,y);
		while(d[x]^d[y]) s[i]|=1ll<<x-1,x=f[x];
		while(x^y) s[i]|=(1ll<<x-1)|(1ll<<y-1),x=f[x],y=f[y];
	}
	int lim=1<<m;
	for(int i=0;i<lim;i++){
		int cnt=0,now=0,tot=0;
		for(int j=0;j<m;j++) if(i>>j&1) cnt++,now|=s[j];
		for(int j=0;j<n;j++) if(now>>j&1) tot++;
		tot=1ll<<(n-1-tot);
		if(cnt&1) ans-=tot;
		else ans+=tot;
	}
	write(ans);
//	write((1ll<<n-1)-ans);
}