Mr. Kitayuta has just bought an undirected graph with n vertices and m edges. The vertices of the graph are numbered from 1 to n. Each edge, namely edge i, has a color ci, connecting vertex ai and bi.

Mr. Kitayuta wants you to process the following q queries.

In the i-th query, he gives you two integers - ui and vi.

Find the number of the colors that satisfy the following condition: the edges of that color connect vertex ui and vertex vi directly or indirectly.

Input

The first line of the input contains space-separated two integers - n and m(2 ≤ n ≤ 105, 1 ≤ m ≤ 105), denoting the number of the vertices and the number of the edges, respectively.

The next m lines contain space-separated three integers - ai, bi(1 ≤ ai < bi ≤ n) and ci(1 ≤ ci ≤ m). Note that there can be multiple edges between two vertices. However, there are no multiple edges of the same color between two vertices, that is, if i ≠ j, (ai, bi, ci) ≠ (aj, bj, cj).

The next line contains a integer- q(1 ≤ q ≤ 105), denoting the number of the queries.

Then follows q lines, containing space-separated two integers - ui and vi(1 ≤ ui, vi ≤ n). It is guaranteed that ui ≠ vi.

Output

For each query, print the answer in a separate line.

Examples

Input

4 5
1 2 1
1 2 2
2 3 1
2 3 3
2 4 3
3
1 2
3 4
1 4


Output

2
1
0


Input

5 7
1 5 1
2 5 1
3 5 1
4 5 1
1 2 2
2 3 2
3 4 2
5
1 5
5 1
2 5
1 5
1 4


Output

1
1
1
1
2

Note

Let's consider the first sample. 

<image> The figure above shows the first sample. 

  * Vertex 1 and vertex 2 are connected by color 1 and 2. 
  * Vertex 3 and vertex 4 are connected by color 3. 
  * Vertex 1 and vertex 4 are not connected by any single color. 
import java.util.Scanner;

/**
 * Created by yakoub on 06/09/15.
 */
public class UnionFind {
    private int[] P;
    private int[] rank;

    public UnionFind(int n) {
        P = new int[n + 1];
        rank = new int[n + 1];
    }

    public void create(int x) {
        P[x] = x;
    }

    public int findSet(int x) {
        if (x != P[x]) P[x] = findSet(P[x]);
        return P[x];
    }

    public void merge(int a, int b) {

        if (P[a] == 0)
            P[a] = a;
        if (P[b] == 0)
            P[b] = b;
        int pa = findSet(a);
        int pb = findSet(b);
        if (rank[pa] > rank[pb])
            P[pb] = pa;
        else
            P[pa] = pb;
        if (rank[pa] == rank[pb])
            rank[pb] = rank[pb] + 1;

    }

    public boolean same(int a, int b) {
        int pa = findSet(a);
        int pb = findSet(b);
        return (pa == pb) && (pa != 0);
    }

    public static void main(String args[]) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        int m = scanner.nextInt();
        UnionFind colors[] = new UnionFind[m + 1];
        for (int i = 0; i < m; i++) {
            int from = scanner.nextInt();
            int to = scanner.nextInt();
            int color = scanner.nextInt();
            if (colors[color] == null)
                colors[color] = new UnionFind(n);
            colors[color].merge(from, to);
        }
        int q = scanner.nextInt();
        for (int i = 0; i < q; i++) {
            int from = scanner.nextInt();
            int to = scanner.nextInt();
            int count = 0;
            for (int j = 1; j <= m; j++) {
                UnionFind set = colors[j];
                if (set != null)
                    if (set.same(from, to))
                        count++;
            }
            System.out.println(count);
        }
    }
}
