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.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.StringTokenizer;

public class B505 {

    static boolean[][][] edges;
    static int n;
    static boolean isOk;
    static boolean[] visited;

    public static void main(String[] args) throws IOException {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
        StringTokenizer st = new StringTokenizer(in.readLine());
        n = Integer.parseInt(st.nextToken());
        int m = Integer.parseInt(st.nextToken());
        edges = new boolean[m][n][n];
        for (int i = 0; i < m; i++) {
            st = new StringTokenizer(in.readLine());
            int a = Integer.parseInt(st.nextToken()) - 1;
            int b = Integer.parseInt(st.nextToken()) - 1;
            int c = Integer.parseInt(st.nextToken()) - 1;
            edges[c][a][b] = edges[c][b][a] = true;
        }
        st = new StringTokenizer(in.readLine());
        int q = Integer.parseInt(st.nextToken());
        for (int i = 0; i < q; i++) {
            st = new StringTokenizer(in.readLine());
            int a = Integer.parseInt(st.nextToken()) - 1;
            int b = Integer.parseInt(st.nextToken()) - 1;
            int cnt = 0;
            for (int color = 0; color < m; color++) {
                isOk = false;
                visited = new boolean[n];
                dfs(a, b, color);
                if (isOk) {
                    cnt++;
                }
            }
            out.println(cnt);
        }
        out.close();
    }

    public static void dfs(int v, int finish, int color) {
        if (v == finish) {
            isOk = true;
            return;
        } else {
            visited[v] = true;
            for (int i = 0; i < n; i++) {
                if (edges[color][v][i] && !visited[i]) {
                    dfs(i, finish, color);
                }
            }
        }
    }

}
