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;
using Int = long long;
int main() {
    int N; cin >> N;
    vector<vector<pair<int, int>>> G(N);
    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);
    }
    int M; cin >> M;
    vector<Int> B(M);
    function<Int(int, int, Int)> recur = [&](int u, int v, Int acc) -> Int {
        if (u == v) return acc;
        for (auto& edge : G[u]) if (~acc & (1LL << edge.second)) {
            auto r = recur(edge.first, v, acc | (1LL << edge.second));
            if (r != 0) return r;
        }
        return 0;
    };
    for (int i = 0; i < M; i++) {
        int u, v; cin >> u >> v; u--, v--;
        B[i] = recur(u, v, 0);
    }
    vector<Int> TWO(N);
    TWO[0] = 1;
    for (int i = 1; i < TWO.size(); i++) TWO[i] = TWO[i - 1] * 2;
    Int ans = 0;
    for (int i = 0; i < (1 << M); i++) {
        Int bits = 0;
        for (int j = 0; j < M; j++) if (i & (1LL << j)) {
            bits |= B[j];
        }
        int mcnt = __builtin_popcountll(i);
        int ncnt = __builtin_popcountll(bits);
        ans += (mcnt % 2 ? -1 : +1) * TWO[N - 1 - ncnt];
    }
    cout << ans << endl;
    return 0;
}