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<algorithm>
#include<vector>
#include<queue>
#include<set>
#include<unordered_set>
#include<unordered_map>
using namespace std;
typedef long long ll;
#define chmax(a,b) a=max(a,b)
#define chmin(a,b) a=min(a,b)
#define all(x) x.begin(),x.end()
#define rep(i,n) for(int i=0;i<n;i++)
#define mod 1000000007
#define mad(a,b) a=(a+b)%mod
#define N 200010
vector<ll> g[55];
ll n,m,a[55],b[55];
ll dist[55][55];
void dfs(ll x,ll root,ll par,ll di){
    dist[root][x]=di;
    for(auto y:g[x])if(y!=par){
	dfs(y,root,x,di+1);
    }
}
ll v[55];
ll dp[(1<<20)];
int main(){
    cin>>n;
    rep(i,n-1){
	cin>>a[i]>>b[i]; a[i]--,b[i]--;
	g[a[i]].push_back(b[i]);
	g[b[i]].push_back(a[i]);
    }
    rep(i,n)dfs(i,i,-1,0);
    rep(i,n)v[i]=0;
    cin>>m;
    rep(i,m){
	ll s,t; cin>>s>>t; s--,t--;
	rep(j,n-1){
	    if(dist[s][a[j]]+dist[a[j]][t]==dist[s][t]&&dist[s][b[j]]+dist[b[j]][t]==dist[s][t]){
		v[j]+=(1LL<<i);
	    }
	}
    }
    rep(i,(i<<m))dp[i]=0;
    dp[0]=1;
    rep(i,n-1)for(int j=(1<<m)-1;j>=0;j--){
	dp[j|v[i]]+=dp[j];
    }
    cout<<dp[(1<<m)-1]<<endl;
}


