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.HashSet;
import java.util.Scanner;
import java.util.Set;

/**
 * Created by kate on 18/01/15.
 */
public class Ketauta {
    static class Edge {
        int a;
        int b;
        int c;

        public Edge(int a, int b, int c) {
            this.a = a;
            this.b = b;
            this.c = c;
        }
    }
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt(); //vertices
        int m = sc.nextInt(); //edges
        Edge[] e = new Edge[m];
        for (int i = 0; i < m; i++) {
            e[i] = new Edge(sc.nextInt(), sc.nextInt(), sc.nextInt());
        }
        int q = sc.nextInt();
        for (int i = 0; i < q; i++) {
            System.out.println(countColors(e, sc.nextInt(), sc.nextInt()));
        }
    }

    private static int countColors(Edge[] edges, int u, int v) {
        Set<Integer> colors = new HashSet<>();

        for (Edge edge : edges) {
            if ((edge.a == u || edge.b == u) && !colors.contains(edge.c)) {
                Set<Edge> path = new HashSet<>();
                path.add(edge);
                int c;
                if (edge.a == v || edge.b == v) {
                    c = edge.c;
                } else {
                    c = crawl(edges, v, edge.c, path);
                }
                if (c > 0)
                    colors.add(c);
            }
        }
        return colors.size();
    }

    private static int crawl(Edge[] edges, int v, int c, Set<Edge> path) {
        for (Edge edge : edges) {
            if (edge.c == c && !path.contains(edge)) {
                if(connected(edge, path)) {
                    path.add(edge);
                    if (edge.a == v || edge.b == v) {
                        return c;
                    } else {
                        return crawl(edges, v, c, path);
                    }
                }
            }
        }
        return -1;
    }

    private static boolean connected(Edge edge, Set<Edge> path) {
        for (Edge e: path){
            if(e.a == edge.a || e.a == edge.b || e.b == edge.a || e.b == edge.b){
                return true;
            }
        }
        return false;
    }
}
