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.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashSet;
import java.util.Iterator;
import java.util.LinkedList;
import java.util.ListIterator;
import java.util.StringTokenizer;
import javafx.util.Pair;

/**
 *
 * @author Public
 */
public class D506 {
    public static int count=0;
    static boolean found=false;
      public static void main(String[] args) throws Exception{
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st=new StringTokenizer(br.readLine());
        int n=Integer.parseInt(st.nextToken());
        int m=Integer.parseInt(st.nextToken());
        LinkedList ad[]=new LinkedList[n+1];
        boolean visited[]=new boolean[n+1];
        HashSet l[]=new HashSet[n+1];
        for(int i=0;i<n+1;i++){
        ad[i]=new LinkedList<Pair<Integer,Integer>>();
        l[i]=new HashSet();
        }
        for(int i=0;i<m;i++){
        StringTokenizer st1=new StringTokenizer(br.readLine());
        int u=Integer.parseInt(st1.nextToken());
        int v=Integer.parseInt(st1.nextToken());
        int c=Integer.parseInt(st1.nextToken());
       
        l[u].add(c);
        l[v].add(c);
        ad[u].add(new Pair(c,v));
        ad[v].add(new Pair(c,u));
        }
            class dfs{
          
  void DFS(int x,int y,int c,boolean[] v){
     v[x]=true;
      if(x==y || found==true){
     found=true;
     return;
     }
  ListIterator itr= ad[x].listIterator();
  while(itr.hasNext()){
      if(found==true)
          break;
  Pair p=(Pair)itr.next();
      if((int)p.getKey()==c && v[(int)p.getValue()]==false){
        DFS((int)p.getValue(),y,c,v);
      }
  }
  }
 }
        
        int q=Integer.parseInt(br.readLine());
        for(int i=0;i<q;i++){
         StringTokenizer st2=new StringTokenizer(br.readLine());
        int x=Integer.parseInt(st2.nextToken());
        int y=Integer.parseInt(st2.nextToken());
        HashSet tx=new HashSet(l[x]);
        tx.retainAll(l[y]);
            Iterator itr=tx.iterator();
            while(itr.hasNext()){
                Arrays.fill(visited,false);
            int color=(int)itr.next();
            new dfs().DFS(x, y, color,visited);
            if(found==true){count++; found=false;}
            }
            System.out.println(count);
        count=0;
        }
      
      
      }
}
