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
//package com.company;
import java.util.*;

public class Main {
    static ArrayList<Edge>[] gr;
    static boolean[] mark;
    static int[] cover;

    static class Edge {
        int to, num;
        Edge(int to, int num) {
            this.to = to;
            this.num = num;
        }
    }

    static boolean DFS(int x, int y, int p) {
        if (x == y) return true;
        for (Edge e : gr[x]) {
            if (e.to == p) continue;
            if (DFS(e.to, y, x)) {
                mark[e.num] = true;
                return true;
            }
        }
        return false;
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        gr = new ArrayList[n];
        for (int i = 0; i < n; i++)
            gr[i] = new ArrayList<>();
        for (int i = 0; i < n - 1; i++)
        {
            int u = sc.nextInt(), v = sc.nextInt();
            u--;
            v--;
            gr[u].add(new Edge(v, i));
            gr[v].add(new Edge(u, i));
        }
        int m = sc.nextInt();
        cover = new int[n];
        for (int i = 0; i < m; i++)
        {
            int u = sc.nextInt(), v = sc.nextInt();
            u--;
            v--;

            mark = new boolean[n];
            for (int j = 0; j < n; j++)
                mark[j] = false;

            DFS(u, v, -1);

            for (int j = 0; j < n - 1; j++)
                if (mark[j])
                    cover[j] |= (1 << i);
        }

        long d[] = new long[1 << m];
        d[0] = 1;

        for (int i = 0; i < n - 1; i++)
            for (int x = (1 << m) - 1; x >= 0; x--) {
                long q = d[x];
                //d[x] += q;
                d[x | cover[i]] += q;
            }

        System.out.println(d[(1 << m) - 1]);
    }
}
