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:
#define HAVE_STRUCT_TIMESPEC
#include<bits/stdc++.h>
using namespace std;
vector<pair<int,int> >v[57];
long long dp[1<<20];
int x[57];
bool dfs(int now,int fa,int root,int pos){//当前结点，父节点，终结点，所在边编号
    if(now==root)
        return 1;
    for(auto it:v[now]){
        if(it.first!=fa&&dfs(it.first,now,root,pos)){
            x[it.second]^=1<<pos;
            return 1;
        }
    }
    return 0;
}
int main(){
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int n;
    cin>>n;
    for(int i=1;i<n;++i){
        int a,b;
        cin>>a>>b;
        v[a].emplace_back(b,i);
        v[b].emplace_back(a,i);
    }
    int m;
    cin>>m;
    for(int i=0;i<m;++i){
        int a,b;
        cin>>a>>b;
        dfs(a,0,b,i);//遍历a和b中间的所有结点，即遍历两点之间所有边
    }
    dp[0]=1;//所有的边全白
    for(int i=1;i<n;++i)//每次添加一条边
        for(int j=(1<<m)-1;j>=0;--j)//状压DP，从后往前是为了避免重复加
            dp[j|x[i]]+=dp[j];//dp【j状态|加入第i条边变黑能满足的状态x[i]】+=dp【第i条边加入之前的状态j】
    cout<<dp[(1<<m)-1];//全状态
    return 0;
}