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.*;
import java.io.*;
 
  
   
    public  class graph{
    
         public static void main (String args[])throws IOException{
             
             BufferedReader ob = new BufferedReader(new InputStreamReader(System.in));
             
            StringTokenizer st = new StringTokenizer(ob.readLine());
         
            int n = Integer.parseInt(st.nextToken());
            int m = Integer.parseInt(st.nextToken());
               DisjointSet []ds = new DisjointSet[m];
               
               for(int i = 0; i < m; i++){
                  ds[i] = new DisjointSet(n); 
               }
               for(int i = 0; i <  m; i++){
                   StringTokenizer st1 = new StringTokenizer(ob.readLine());
                   int a = Integer.parseInt(st1.nextToken())-1;
                   int b = Integer.parseInt(st1.nextToken())-1;
                   int c = Integer.parseInt(st1.nextToken())-1;
                   ds[c].union(a,b);
               }
               int q = Integer.parseInt(ob.readLine());
               for(int i = 0; i < q; i++){
                   StringTokenizer st2 = new StringTokenizer(ob.readLine());
                   int ans = 0;
                   int u = Integer.parseInt(st2.nextToken())-1;
                   int v = Integer.parseInt(st2.nextToken())-1;
                   for(int j = 0; j < m; j++){
                       if(ds[j].find(u) == ds[j].find(v)){
                           ans++;
                       }
                   }
                   System.out.println(ans);
               }
         }
         
         
         
         
         
         public static class DisjointSet{
             int []map;
             
             public DisjointSet(int n){
                 map = new int[n+1];
                 Arrays.fill(map,-1);
             }
             public int find(int x){
                 if(map[x] < 0){
                     return x;
                 }
                 return map[x]= find(map[x]);
             }
             
             public  void union(int a , int b){
                 int roota = find(a);
                 int rootb = find(b);
                 
                if(roota == rootb){
                    return;
                }else if(map[roota] < map[rootb]){
                    map[roota] += map[rootb];
                    map[rootb] = roota;
                }else{
                    map[rootb] += map[roota];
                map[roota] = rootb;
                }
             }
         }
      }