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 int long long
using namespace std;
constexpr int mod=1e9+7;
struct edge{int to,idx;};
vector<edge>G[50];
int pre[50];
void dfs(int s,int p){
    for(int i=0;i<G[s].size();i++){
        if(p!=G[s][i].to){
            pre[G[s][i].to]=pre[s]^1ll<<G[s][i].idx;
            dfs(G[s][i].to,s);
        }
    }
}
signed main(){
    int n,a,b,m,u,v,egs[20];
    cin>>n;
    for(int i=0;i<n-1;i++){
        cin>>a>>b;
        a--,b--;
        G[a].push_back(edge{b,i});
        G[b].push_back(edge{a,i});
    }
    cin>>m;
    for(int i=0;i<m;i++){
        cin>>u>>v;
        u--,v--;
        pre[u]=(1ll<<n-1)-1;
        dfs(u,-1);
        egs[i]=pre[v];
    }
    int ans=0;
    for(int s=1;s<1ll<<m;s++){
        int tmp=(1ll<<n-1)-1;
        for(int i=0;i<m;i++)
            if(s&1ll<<i)
                tmp&=egs[i];
        int c=__builtin_popcountll(tmp);
        if(__builtin_parityll(s))
            ans+=1ll<<c;
        else
            ans-=1ll<<c;
    }
    cout<<(1ll<<n-1)-ans<<endl;
}
