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;

int n;
vector<int> g[55];
int dep[55],fa[55];
int m,U[55],V[55];
bool fix[55];
int tot;
long long ans;

void dfs(int x,int p){
    for(int &y:g[x])if(y!=p){
        fa[y]=x;
        dep[y]=dep[x]+1;
        dfs(y,x);
    }
}

int main(){

    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    cin>>n;
    for(int i=1;i<n;i++){
        int u,v;
        cin>>u>>v;
        g[u].emplace_back(v);
        g[v].emplace_back(u);
    }
    dfs(1,-1);
    cin>>m;
    for(int i=0;i<m;i++)cin>>U[i]>>V[i];
    for(int s=0;s<1<<m;s++){
        memset(fix,0,sizeof(fix));
        tot=n-1;
        for(int i=0;i<m;i++)if(s>>i&1){
            int u=U[i],v=V[i];
            if(dep[u]<dep[v])swap(u,v);
            while(dep[u]>dep[v])fix[u]=1,u=fa[u];
            while(u!=v)fix[u]=1,fix[v]=1,u=fa[u],v=fa[v];
        }
        for(int i=2;i<=n;i++)tot-=fix[i];
        if(__builtin_popcount(s)&1)ans-=1ll<<tot;
        else ans+=1ll<<tot;
    }
    cout<<ans<<endl;

    return 0;
}