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>
using namespace std;

int F[55];
long long dp[2][(1 << 20)];
vector< pair<int, int> > X[55];

int dfs(int i, int par, int target, int curr)
{
    if(i == target)
        return 1;
    
    int ans = 0;
    
    for(auto x : X[i])
        if(x.first != par)
        {
            int y = dfs(x.first, i, target, curr);
            ans |= y;

            if(y == 1)
                F[x.second] |= curr;
        }
    
    return ans;
}

int main()
{
    ios::sync_with_stdio(0);
    cin.tie(NULL), cout.tie(NULL);

    int N;
    cin >> N;

    for(int i=1; i<=N-1; i++)
    {
        int a, b;
        cin >> a >> b;

        X[a].push_back({b, i});
        X[b].push_back({a, i});
    }

    int M;
    cin >> M;

    for(int i=0; i<M; i++)
    {
        int u, v;
        cin >> u >> v;

        dfs(u, 0, v, (1 << i));
    }

    int prev = 0, curr = 1, goal = (1 << M) - 1;
    dp[0][0] = 1;

    for(int i=1; i<=N-1; i++, swap(prev, curr))
    {
        for(int j=0; j<=goal; j++)
            dp[curr][j] = dp[prev][j];
        
        for(int j=0; j<=goal; j++)
            dp[curr][(j | F[i])] += dp[prev][j];
    }
    
    cout << dp[prev][goal] << "\n";
    return 0;
}
