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 lint = long long int;
#define bit(n) (1LL<<(n))

vector<pair<lint, lint>> edges[51];
bitset<30> bs[51]; // i -> {mask}
lint dp[51][bit(21)];

bool dfs(lint node, lint goal, lint from, lint label) {
    bool past = (node == goal);

    for(auto elem : edges[node]) {
        lint next, index;
        tie(next, index) = elem;

        if(next == from)  continue;
        if(dfs(next, goal, node, label)) {
            past = true;
            bs[index].set(label);
        }
    }
    return past;
}

int main() {
    lint n;
    cin >> n;

    for(lint i=0; i<n-1; i++) {
        lint a, b;
        cin >> a >> b;
        a--, b--;
        edges[a].emplace_back(b, i);
        edges[b].emplace_back(a, i);
    }

    lint m;
    cin >> m;

    // 1. 辺i -> 制約{mask[i]}
    for(lint i=0; i<m; i++) {
        lint u, v;
        cin >> u >> v;
        u--, v--;
        dfs(u, v, -1, i);
    }

    // 2. bitDP
    // dp[i][mask] := 辺[0, i)を塗り分ける場合のうち、制約{mask}を満たすものの個数
    dp[0][0] = 1;
    for(lint i=0; i<n-1; i++)
    for(lint mask=0; mask<bit(m); mask++) {
        dp[i+1][mask] += dp[i][mask];
        dp[i+1][mask | bs[i].to_ullong()] += dp[i][mask];
    }

    cout << dp[n-1][bit(m)-1] << endl;
}