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. 
/*
PROG: b
LANG: JAVA
*/
import java.util.*;
import java.io.*;

public class b {
  final int N = 111;
  boolean doit(int p, int t, boolean[][] g, boolean[] v) {
    if(p == t) return true; 
    if(v[p]) return false;
    v[p] = true;
    boolean r = false;
    for(int i = 0; i < g.length; ++i) {
      if(g[p][i]) 
        r |= doit(i, t, g, v); 
    }
    return r;
  }
  private void solve() throws IOException {
    int n = nextInt(), m = nextInt();
    boolean[][][] g = new boolean[N][N][N];

    for(int i = 0; i < m; ++i) {
      int a = nextInt() - 1, b = nextInt() - 1, c = nextInt() - 1;
      g[c][a][b] = true;
      g[c][b][a] = true;
    }
    
    int q = nextInt();
    for(int i = 0; i < q; ++i) {
      int a = nextInt() - 1, b = nextInt() - 1, r = 0; 
      for(int j = 0; j < m; ++j) {
        boolean[] v = new boolean[n];
        if(doit(a, b, g[j], v)) {
          ++r;
        }
      }
      out.println(r);
    }
  }

  static void debug(Object...o) {
    System.err.println(Arrays.deepToString(o));
  }

  BufferedReader br;
  StringTokenizer st;
  PrintWriter out = new PrintWriter(System.out);

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

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

  double nextDouble() throws IOException {
    return Double.parseDouble(next());
  }

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

  private void run() throws IOException {
    //br = new BufferedReader(new FileReader("b.in"));
    br = new BufferedReader(new InputStreamReader(System.in));
    solve();
    out.close();
  }

  public static void main(String[] args) throws IOException {
    new b().run();
  }
}
