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:
#include <bits/stdc++.h>
using namespace std;

#define MAX_N 51

int n ,m;
int edge_id[MAX_N][MAX_N];
vector <int> adj[MAX_N] ,in_res[MAX_N];

vector <int> get_path(int a ,int b){
    queue <int> q{{a}};
    vector <int> from(n+1);
    while(!q.empty()){
        int u = q.front(); q.pop();
        if(u == b){
            vector <int> path;
            for(int v=u; v!=a; v=from[v])
                path.push_back(edge_id[v][from[v]]);
            return path;
        }
        for(int&v : adj[u])
            if(from[u] != v)
                q.push(v) ,from[v] = u;
    }
}
long long dp[MAX_N][1<<20];
long long solve(int e ,int b){
    if(e == n-1)
        return b == (1<<m)-1;
    auto&ret = dp[e][b];
    if(~ret)
        return ret;

    ret = solve(e+1 ,b);
    for(int&r : in_res[e])
        b |= (1<<r);
    return ret += solve(e+1 ,b);
}

int main()
{
    scanf("%d",&n);
    for(int a,b,i=1; i<n; i++){
        scanf("%d%d",&a,&b);
        edge_id[a][b] = i-1;
        edge_id[b][a] = i-1;
        adj[a].push_back(b);
        adj[b].push_back(a);
    }

    scanf("%d",&m);
    for(int a,b,i=0; i<m; i++){
        scanf("%d%d",&a,&b);
        for(int&e : get_path(a ,b))
            in_res[e].push_back(i);
    }

    memset(dp ,-1 ,sizeof dp);
    printf("%lld\n",solve(0 ,0));
}