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>

#define FILE
#define fr first
#define se second

using namespace std;

const long long N = 5e1 + 7;
const long long inf = 1e9 + 7;
const long long mod = 1e9 + 7;

int n;
int m;
int cover[N];
vector<pair<int, int>> g[N];
bool dfs(int v, int fi, int p, int idx){
    if(v == fi){
        return true;
    }
    for(auto to: g[v]){
        if(to.fr == p){
            continue;
        }
        if(dfs(to.fr, fi, v, idx)){
            cover[to.se] |= (1 << idx);
            return true;
        }
    }
    return false;
}
int main()
{
    #ifdef FILEs
        freopen("input.txt", "r", stdin);
        /// freopen("output.txt", "w", stdout);
    #endif

    ios_base::sync_with_stdio(false);

    cin >> n;
    for(int i = 0; i < n - 1; i ++){
        int x, y;
        cin >> x >> y;
        g[x].push_back({y, i});
        g[y].push_back({x, i});
    }
    cin >> m;
    for(int i = 0; i < m; i ++){
        int x, y;
        cin >> x >> y;
        dfs(x, y, -1, i);
    }
    vector<long long> d(1 << m);
    d[0] = 1;
    for(int i = 0; i < n - 1; i ++){
        for(int mask = (1 << m) - 1; mask >= 0; mask --){
            d[mask | cover[i]] += d[mask];
        }
    }
    cout << d[(1 << m) - 1] << "\n";
}

