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.awt.List;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.StringTokenizer;

public class Codeforces505B {
    private static InputReader in;
    private static PrintWriter out;
    static ArrayList<Integer>[][] edges; //edges[i][c] gives the indices of the vertices connected to i with color c
    static int n;
    static int m;
    static boolean[] visited;
    static int c;
    static int counts;
    public static void dfs(int node){
        visited[node]=true;
        for(int node2: edges[node][c]){
            if(!visited[node2]) dfs(node2);
        }
    }
    
    public static void query(int u, int v, int ci){
        visited= new boolean[n+1];
        c=ci;
        dfs(u);
        if(visited[u] && visited[v]){
            counts++; 
        }
    }
    
    public static void main(String[] args) {
        in = new InputReader(System.in);
        out = new PrintWriter(System.out, true);
        n=in.nextInt();
        m= in.nextInt();
        edges= new ArrayList[n+1][m+1];
        for(int i=1; i<=n; i++){
            for(int j=1; j<=m; j++){
                edges[i][j]= new ArrayList<Integer>();
            }
        }
        for(int count=0; count<m; count++){
            int a= in.nextInt();
            int b= in.nextInt();
            int c= in.nextInt();
            edges[a][c].add(b);
            edges[b][c].add(a);
        }
        int q= in.nextInt();
        for(int count=0; count<q; count++){
            int u= in.nextInt();
            int v= in.nextInt();
            counts=0;
            for(int c=1; c<=m; c++){
                query(u,v,c);
            }
            out.println(counts);
        }
    }
    
    
    
    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
          reader = new BufferedReader(new InputStreamReader(stream), 32768);
          tokenizer = null;
        }

        public String next() {
          while (tokenizer == null || !tokenizer.hasMoreTokens()) {
            try {
              tokenizer = new StringTokenizer(reader.readLine());
            } catch (IOException e) {
              throw new RuntimeException(e);
            }
          }
          return tokenizer.nextToken();
        }

        public int nextInt() {
          return Integer.parseInt(next());
        }
    }
}
