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>
#define mp make_pair
using namespace std;
typedef long long ll;
double ans;
vector<pair<int,int> > G[100];
int n,m;
ll sta,St[100];
ll dfs(int u,int fa,int ed,int eg){
    if(u==ed)return 1ll<<eg;
    if(G[u].size()==1&&~fa){return 0;}
    ll ret=0;
    for(auto v: G[u])if(v.first!=fa){
        ret|=dfs(v.first,u,ed,v.second);
        if(ret&&eg>=0)ret|=1ll<<eg;
        //cerr<<u<<" "<<v.first<<" "<<ret<<" "<<eg<<endl;
    }
    return ret;
}
void solve(){
    scanf("%d",&n);
    for(int i=1,u,v;i<n;++i){
        scanf("%d%d",&u,&v);
        G[u].push_back(mp(v,i-1));
        G[v].push_back(mp(u,i-1));
    }
    scanf("%d",&m);
    for(int i=1,u,v;i<=m;++i){
        scanf("%d%d",&u,&v);
        St[i-1]=dfs(u,-1,v,-1);
        //cerr<<"State:: "<<St[i-1]<<endl;
    }
    for(int s=0,c,S=1<<m,cnt;s<S;++s){
        c=__builtin_popcount(s);
        sta=0;
        for(int i=0;i<m;++i)
            if(s&(1<<i))sta|=St[i];
        cnt=0;
        for(int i=0;i<n;++i)
            if(sta&(1ll<<i))++cnt;
        ans+=pow(2,n-1-cnt)*pow(-1,c);
    }
    printf("%.0lf\n",ans);
}
int main(){
    solve();

    return 0;
}