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;

using pp = pair<int, int>;
int N, M;
vector<vector<pp>> G;
vector<int> u, v;

bool dfs(int u, int p, int v, long long &path) {
    if(u == v) {
        path = 0;
        return true;
    }
    for(auto e : G[u]) {
        if(e.first == p) continue;
        if(dfs(e.first, u, v, path)) {
            path |= (1LL << e.second);
            return true;
        }
    }
    return false;
}

long long solve() {
    vector<long long> path(M, 0);
    for(int i=0; i<M; ++i) dfs(u[i], -1, v[i], path[i]);
    
    long long res = 0;
    for(int k=0; k<(1<<M); ++k) {
        long long vacant = 0;
        for(int i=0; i<M; ++i) {
            if(k & (1 << i)) vacant |= path[i];
        }
        long long cnt = N - 1 - __builtin_popcountll(vacant);
        if(__builtin_popcount(k) % 2 == 0) res += (1LL << cnt);
        else res -= (1LL << cnt);
    }
    return res;
}

int main() {
    cin >> N;
    G.resize(N);
    for(int i=0; i+1<N; ++i) {
        int a, b; cin >> a >> b;
        --a; --b;
        G[a].emplace_back(b, i);
        G[b].emplace_back(a, i);
    }
    cin >> M;
    u.resize(M); v.resize(M);
    for(int i=0; i<M; ++i) {
        cin >> u[i] >> v[i];
        --u[i]; --v[i];
    }
    cout << solve() << '\n';
}