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.ArrayList;
import java.util.List;
import java.util.Scanner;
import java.util.Set;
import java.util.TreeSet;

/**
 *
 * @author Meo Nam
 */
public class UVA {
    static List<ArrayList<Edge>> list = new ArrayList(100);
    static List<Set<Integer>> color = new ArrayList(100);
    static boolean checked = false;
    static class Edge {
        public int x;
        public int y;
        public int c;
        
        public Edge(int x, int y, int c) {
            this.x = x;
            this.y = y;
            this.c = c;
        }
    }
    
    static void dfs(int u, boolean[] isVisited, int color, int v) {
        isVisited[u] = true;
        if (u == v )  {
            checked = true;
        }
        
        for (Edge e : list.get(u)) {
            if (!isVisited[e.y] && e.c == color) {
                dfs(e.y, isVisited, color, v);
            }
        }
        //return false;
    }
    
    
    public static void main(String arg[]) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();
        for (int i=0; i<=n+1; i++) {
           list.add(i,new ArrayList<Edge>() );
           color.add(i,new TreeSet<Integer>());
        }
        //list.add(new ArrayList<Edge>());
        //color.add(new ArrayList<Integer>());
        for (int i=0; i<m; i++) {
            int a = sc.nextInt();
            int b = sc.nextInt();
            int c = sc.nextInt();
            
            list.get(a).add(new Edge(a,b,c));
            list.get(b).add(new Edge(b,a,c));
            
            color.get(a).add(c);
            color.get(b).add(c);
        }
        int q = sc.nextInt();
        
        boolean [] isVisited = new boolean[n+1];
        for (int k=0; k<=n; k++) {
               isVisited[k] = false;
        }
        for (int i=0; i<q; i++) {
            int u = sc.nextInt();
            int v = sc.nextInt();
           
            int count = 0;
            for (int c : color.get(u)) {
                if (color.get(v).contains(c)) {
                    checked = false;
                    for (int k=1; k<=n; k++) {
                        isVisited[k] = false;
                    }
                     dfs(u, isVisited, c, v);
                     if (checked)    
                        count++;
                    
                }
            }
            System.out.println(count);
        }
    }
}