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:
/* by Natsu Kinmoe */
#include <bits/stdc++.h>
using namespace std;

#define all(x) (x).begin(),(x).end()
#define loop(i,n) for(int i=0;i<(n);i++)
#define cont(i,n) for(int i=1;i<=(n);i++)
#define circ(i,a,b) for(int i=(a);i<=(b);i++)
#define range(i,a,b,c) for(int i=(a);((c)>0?i<=(b):i>=(b));i+=(c))
#define foreach(it,v) for(__typeof((v).begin()) it=(v).begin(),_e_D_=(v).end();it!=_e_D_;it++)
#define y0 y0O0OO00OO0OO0OO0OOO00OO0OO0O0O000OO0
#define y1 y1II11II11III11I1III11II111IIII1II1I1
#define pub push_back
#define pob pop_back
#define mak make_pair

typedef long long ll;
typedef long double lf;
const int Inf=0x3f3f3f3f;
const ll INF=0x3f3f3f3f3f3f3f3fll;
/* Source code starts here */

int n,m;
int nei[55][55];
int hv[55];
ll dp[1<<20];

bool dfs(int now,int lst,int to,int id){
	if(now==to)return 1;
	cont(i,n)if(i!=lst&&nei[now][i]&&dfs(i,now,to,id))return hv[nei[now][i]]|=(1<<id),1;
	return 0;
}
 
int main(){
	scanf("%d",&n);
	cont(i,n-1){
		int a,b;scanf("%d%d",&a,&b);
		nei[a][b]=nei[b][a]=i;
	}
	scanf("%d",&m);
	loop(i,m){
		int a,b;scanf("%d%d",&a,&b);
		dfs(a,0,b,i);
	}
	dp[0]=1;
	cont(i,n-1)range(j,(1<<m)-1,0,-1)dp[j|hv[i]]+=dp[j];
	printf("%lld\n",dp[(1<<m)-1]);
	return 0;
}