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 REP(i,n) for(int i=0,i##_len=int(n);i<i##_len;++i)
#define rep(i,a,b) for(int i=int(a);i<int(b);++i)
#define All(x) (x).begin(),(x).end()
#define rAll(x) (x).rbegin(),(x).rend()
using namespace std;
using ll = long long;
vector<vector<pair<int,int>>> tree;
vector<int> flag;
bool dfs(int u,int pre,int g,int num){
    if(u==g) return true;
    for(auto e:tree[u]){
        if(e.first==pre) continue;
        if(dfs(e.first,u,g,num)){
            flag[e.second]=flag[e.second]|(1<<num);
            return true;
        }
    }
    return false;
}

int main(){
    int N;cin>>N;
    tree.resize(N);
    REP(i,N-1){
        int a,b;
        cin>>a>>b;
        a--;b--;
        tree[a].push_back({b,i});
        tree[b].push_back({a,i});
    }
    int M;cin>>M;
    flag.resize(N-1,0);
    REP(i,M){
        int u,v;
        cin>>u>>v;
        u--;v--;
        dfs(u,-1,v,i);
    }
    vector<ll> dp(1<<M);
    dp[0]=1;
    REP(i,N-1) for(int j=(1<<M)-1;j>=0;--j){
        dp[j|flag[i]]+=dp[j];
    }
    cout<<dp[(1<<M)-1]<<endl;
}
