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;
vector<vector<int> >g(52);
int edgenum[52][52], edgemask[52], par[52];
void dfs(int v, int p){
    par[v] = p;
    for(int i = 0; i < g[v].size(); i++){
        int to = g[v][i];
        if(to == p){
            continue;
        }
        dfs(to, v);
    }
}
void calc(int src, int des, int num){
    dfs(src, 0);
    int v = des;
    while(v != src){
        int u = par[v];
        edgemask[edgenum[v][u]] |= (1 << num);
        v = u;
    }
}
int n, m;
long long dp[51][(1 << 20) + 2];
long long solve(int pos, int mask){
    if(pos == n){
        if(__builtin_popcount(mask) == m){
            return 1;
        }
        return 0;
    }
    if(dp[pos][mask] != -1){
        return dp[pos][mask];
    }
    dp[pos][mask] = solve(pos + 1, mask) + solve(pos + 1, mask | edgemask[pos]);
    return dp[pos][mask];
}
int main(){
    cin >> n;
    for(int i = 1; i < n; i++){
        int u, v;
        cin >> u >> v;
        g[u].push_back(v);
        g[v].push_back(u);
        edgenum[u][v] = i;
        edgenum[v][u] = i;
    }
    cin >> m;
    for(int i = 0; i < m; i++){
        int u, v;
        cin >> u >> v;
        calc(u, v, i);
    }
    memset(dp, -1, sizeof(dp));
    cout << solve(1, 0);
    return 0;
}