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 <iostream>
#include <bits/stdc++.h>
#define ll long long

using namespace std;
vector < pair < ll , ll > > v[60];
ll Edges[60],M,dp[60][(1 << 20) + 5],n;
set < ll > st;

void DFS(ll x , ll p , ll y , ll j , ll idx){

    if(x == y){
        for(auto it : st)  Edges[it]|=(1 << j);
        return;
    }
    for(auto it : v[x]){
        if(it.first != p) {
            st.insert(it.second);
            DFS(it.first , x , y , j , it.second);
        }
    }
    st.erase(idx);
}

ll Rec(int i , ll Mask){

    if(i == n)  return (Mask + 1 == (1 << M));

    if(dp[i][Mask] != -1)  return dp[i][Mask];

    ll Res = 0;
    Res = Rec(i + 1 , Mask|Edges[i]) + Rec(i + 1,Mask);
    return dp[i][Mask] = Res;
}

int main()
{
    memset(dp , -1 , sizeof(dp));
    cin >> n;
    for(ll i = 1;i<n;i++){
        ll x,y;  cin >> x >> y;
        v[x].push_back({y , i});
        v[y].push_back({x , i});
    }
    cin >> M;
    for(ll i = 0;i<M;i++){
        ll x,y;  cin >> x >> y;
        DFS(x , -1 , y , i , -1);
        st.clear();
    }
    cout << Rec(1 , 0);
    return 0;
}
