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.LinkedList;
import java.util.Queue;
import java.util.StringTokenizer;

public class KitayutasColorfulGraph {

  static BufferedReader in;
  static PrintWriter out;
  static StringTokenizer tok;

  static void solve() throws Exception {
    int n = nextInt(), m = nextInt();
    boolean[][][] g = new boolean[m][n][n];
    for (int i = 0; i < m; i++) {
      int a = nextInt(), b = nextInt(), c = nextInt();
      g[c - 1][a - 1][b - 1] = true;
      g[c - 1][b - 1][a - 1] = true;
    }
    for (int i = 0; i < m; i++)
      for (int j = 0; j < n; j++) {
        Queue<Integer> queue = new LinkedList<>();
        for (int k = 0; k < n; k++)
          if (k != j && g[i][j][k])
            queue.offer(k);
        while (!queue.isEmpty()) {
          int t = queue.poll();
          for (int k = 0; k < n; k++)
            if (k != j && !g[i][j][k] && g[i][t][k]) {
              g[i][j][k] = true;
              g[i][k][j] = true;
              queue.offer(k);
            }
        }
      }
    int q = nextInt();
    for (int i = 0; i < q; i++) {
      int a = nextInt() - 1;
      int b = nextInt() - 1;
      int v1 = Math.min(a, b);
      int v2 = Math.max(a, b);
      int c = 0;
      for (int j = 0; j < m; j++)
        if (g[j][v1][v2])
          c++;
      out.println(c);
    }
  }

  static void dfs(int i, int c) {

  }

  public static void main(String args[]) {
    try {
      in = new BufferedReader(new InputStreamReader(System.in));
      out = new PrintWriter(new OutputStreamWriter(System.out));
      solve();
      in.close();
      out.close();
    } catch (Throwable e) {
      e.printStackTrace();
      System.exit(1);
    }
  }

  static int nextInt() throws IOException {
    return Integer.parseInt(next());
  }

  static int[] nextIntArray(int len, int start) throws IOException {
    int[] a = new int[len];
    for (int i = start; i < len; i++)
      a[i] = nextInt();
    return a;
  }

  static long nextLong() throws IOException {
    return Long.parseLong(next());
  }

  static long[] nextLongArray(int len, int start) throws IOException {
    long[] a = new long[len];
    for (int i = start; i < len; i++)
      a[i] = nextLong();
    return a;
  }

  static String next() throws IOException {
    while (tok == null || !tok.hasMoreTokens()) {
      tok = new StringTokenizer(in.readLine());
    }
    return tok.nextToken();
  }

}
