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.*;
import java.util.*;

public class multiGraphs {
    boolean reach;
    public static void main(String args[])throws Exception {
        multiGraphs obj = new multiGraphs();
        obj.solve();
    }
    void solve() throws Exception{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer sz = new StringTokenizer(br.readLine());
        int n = Integer.parseInt(sz.nextToken());
        int m = Integer.parseInt(sz.nextToken());
        String[] ar1 = new String[m];
        for(int i = 0; i < m; i++)
            ar1[i] = br.readLine();
        int q = Integer.parseInt(br.readLine());
        LinkedList<Integer>[] graphar[] = new LinkedList[100][n];
        for(int i = 0; i < 100; i++) {
            LinkedList<Integer>[] graph = graphar[i];
            for(int j = 0; j < n; j++)
                graph[j] = new LinkedList<Integer>();
        }
        for(String str : ar1) {
            sz = new StringTokenizer(str);
            int v1 = Integer.parseInt(sz.nextToken())-1;
            int v2 = Integer.parseInt(sz.nextToken())-1;
            int c = Integer.parseInt(sz.nextToken())-1;
            LinkedList<Integer>[] graph = graphar[c];
            graph[v1].add(v2);
            graph[v2].add(v1);
        }
        int[][] dp = new int[n][n];
        for(int inar[] : dp)
            Arrays.fill(inar, -1);
        for(int i = 0; i < q; i++) {
            sz = new StringTokenizer(br.readLine());
            int v1 = Integer.parseInt(sz.nextToken())-1;
            int v2 = Integer.parseInt(sz.nextToken())-1;
            //if(dp[v1][v2] == -1) {
                int count = 0;
                for(int j = 0; j < 100; j++) {
                    LinkedList<Integer>[] graph = graphar[j];
                    count += dfs(graph, v1, v2, n);
                }
            //  dp[v1][v2] = count;
            //}
            System.out.println(count);
            //System.out.println(dp[v1][v2]);
        }
    }
    int dfs(LinkedList<Integer>[] graph, int s, int d, int n) {
        boolean[] visited = new boolean[n];
        reach = false;
        visit(visited, graph, s, d);
        if(reach)
            return 1;
        return 0;
    }
    void visit(boolean v[], LinkedList<Integer>[] g, int u, int d) {
        v[u] = true;
        if(u == d) {
            reach = true;
            return;
        }
        for(Integer nei : g[u]) {
            if(reach)
                break;
            if(!v[nei])
                visit(v, g, nei, d);
        }
        v[u] = false;
    }
    
} 