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;

typedef long long ll;
typedef pair <int,int> pii;

const int N = 55;

int n, m;
ll Pass[N];
vector <pii> Adj[N];

bool DFS(int x, int y, int i, int p = 0)
{
    if(x == y)  return true;

    for(int k = 0; k < Adj[x].size(); k++)
    {
        int z = Adj[x][k].first;
        int j = Adj[x][k].second;

        if(z == p)  continue;

        Pass[i] |= (1LL << j);

        if(DFS(z, y, i, x)) return true;

        Pass[i] &= ~(1LL << j);
    }

    return false;
}

ll Rec(int i = 0, int k = 0, ll Mask = 0)
{
    if(i == m)
    {
        ll Res = 1LL << ((ll)n - 1 - __builtin_popcountll(Mask));

        return k & 1 ? -Res : +Res;
    }

    return Rec(i + 1, k, Mask) + Rec(i + 1, k + 1, Mask | Pass[i]);
}

int main()
{
    cin >> n;

    for(int i = 1; i < n; i++)
    {
        int x, y;   scanf("%d%d", &x, &y);

        Adj[x].emplace_back(y, i);
        Adj[y].emplace_back(x, i);
    }

    cin >> m;

    for(int i = 0; i < m; i++)
    {
        int x, y;
        scanf("%d%d", &x, &y);
        DFS(x, y, i);
    }

    cout << Rec() << endl;
}