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;
#define all(x) x.begin(), x.end()
#define x first
#define y second
using ll = long long;
#define int ll

const int N = 64;
int n;
vector<pair<int, int>> g[N];
int b[N];

bool dfs(int v, int p, int aim, int bit) {
    if (v == aim) {
        return true;
    }
    for (auto u : g[v]) {
        if (u.x == p) {
            continue;
        }
        if (dfs(u.x, v, aim, bit)) {
            b[u.y] |= bit;
            return true;
        }
    }
    return false;
}

int dp[1 << 21];

signed main() {
#ifdef LC
    assert(freopen("input.txt", "r", stdin));
#endif // LC
    ios::sync_with_stdio(0);
    cin.tie(0);

    cin >> n;
    for (int i = 0; i < n - 1; ++i) {
        int v, u;
        cin >> v >> u;
        g[v].emplace_back(u, i);
        g[u].emplace_back(v, i);
    }
    int m;
    cin >> m;
    for (int i = 0; i < m; ++i) {
        int v, u;
        cin >> v >> u;
        dfs(v, v, u, 1 << i);
    }
    dp[0] = 1;
    for (int i = 0; i < n - 1; ++i) {
        for (int v = (1 << m) - 1; v >= 0; --v) {
            dp[v | b[i]] += dp[v];
        }
    }
    cout << dp[(1 << m) - 1] << endl;
    return 0;
}