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;

int n, m;
vector<pair<int, int>> g[55];

bool dfs (int v, int p, int t, long long &val) {
    for (const auto &e : g[v]) if (e.first != p) {
        int u, id; tie(u, id) = e;
        if (u == t) { val |= (1LL << id); return true; }
        else if (dfs(u, v, t, val)) {
            val |= (1LL << id); return true;
        }
    }
    return false;
}

int main() {
    cin >> n;
    vector<pair<int, int>> dat;
    for (int i = 0; i < n - 1; i++) {
        int a, b; cin >> a >> b; a--; b--;
        g[a].emplace_back(b, i);
        g[b].emplace_back(a, i);
    }
    cin >> m;
    for (int i = 0; i < m; i++) {
        int a, b; cin >> a >> b; a--; b--;
        dat.emplace_back(a, b);
    }
    
    vector<long long> edge(m, 0);
    for (int i = 0; i < m; i++) {
        dfs(dat[i].first, -1, dat[i].second, edge[i]);
    }
    
    long long ans = (1LL << (n - 1));
    
    for (int s = 1; s < (1<<m); s++) {
        int k = 0;
        long long val = 0;
        for (int i = 0; i < m; i++) {
            if (s & (1 << i)) { val |= edge[i]; k++; }
        }
        
        int res = n - 1 - __builtin_popcountll(val);
        if (k & 1) ans -= (1LL << res);
        else ans += (1LL << res);
    }
    
    cout << ans << '\n';
    
    return 0;
}