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;

struct Edge {
    int to, id;
    Edge(int to, int id) : to(to), id(id) {}
};

vector<Edge> g[55];

vector<int> es;

bool dfs(int v, int tv, int p = -1) {
    if(v == tv) return true;
    for(Edge e : g[v]) {
        if(e.to == p) continue;
        if(dfs(e.to, tv, v)) {
            es.push_back(e.id);
            return true;
        }
    }
    return false;
}

int main()
{
    int n;
    cin >> 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<long long> eset(m);
    for(int i = 0; i < m; i++) {
        int a, b;
        cin >> a >> b;
        a--;
        b--;
        es = vector<int>();
        dfs(a, b);
        for(auto&& e : es) {
            eset[i] |= (1LL << e);
        }
    }
    long long ans = 0;
    for(int i = 0; i < (1 << m); i++) {
        long long mask = 0;
        for(int j = 0; j < m; j++) {
            if(i >> j & 1) mask |= eset[j];
        }
        int white = __builtin_popcountll(mask);
        long long now = 1LL << (n - 1 - white);
        if(__builtin_popcountll(i) % 2 == 0) ans += now;
        else ans -= now;
    }
    cout << ans << endl;
    return 0;
}