Mr. Kitayuta has just bought an undirected graph consisting of 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 ≤ 100, 1 ≤ m ≤ 100), 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 ≤ 100), 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. 
2 1
1 2 1
1
1 2
 2 3
1 2 3
1 2 2
1 2 1
1
1 2
 2 5
1 2 1
1 2 2
1 2 3
1 2 4
1 2 5
1
1 2

1
 3
 5

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

public class CF286B {
	private FastScanner in;
	private PrintWriter out;

	public void solve() throws IOException {
		int n = in.nextInt();
		int m = in.nextInt();
		HashMap<Integer, ArrayList<HashSet<Integer>>> graph = new HashMap<Integer, ArrayList<HashSet<Integer>>>();
		for (int i = 1; i <= n; i++){
			ArrayList<HashSet<Integer>> x = new ArrayList<HashSet<Integer>>();
			for (int j = 1; j <= n*2; j++){
				HashSet<Integer> y = new HashSet<Integer>();
				x.add(y);
			}
			graph.put(i, x);
		}
		for (int i = 0; i < m; i++){
			int a = in.nextInt();
			int b = in.nextInt();
			int c = in.nextInt();
			graph.get(a).get(b).add(c);
			graph.get(b).get(a).add(c);
			for (int k = 0; k < 100; k++){
				for (int j = 1; j <= n; j++){
					if (graph.get(a).get(j).contains(c)){
						graph.get(j).get(b).add(c);
					}
					if (graph.get(b).get(j).contains(c)){
						graph.get(j).get(a).add(c);
					}
					if (graph.get(j).get(a).contains(c)){
						graph.get(b).get(j).add(c);
					}
					if (graph.get(j).get(b).contains(c)){
						graph.get(a).get(j).add(c);
					}
				}
			}
		}
		int q = in.nextInt();
		for (int i = 0; i < q; i++){
			int u = in.nextInt();
			int v = in.nextInt();
			out.println(graph.get(u).get(v).size());
		}
	}

	public static void main(String[] args) {
		new CF286B().run();
	}

	public void run() {
		try {
			in = new FastScanner();
			out = new PrintWriter(System.out);
			solve();
			out.close();
		} catch (IOException e) {
			e.printStackTrace();
		}
	}

	private class FastScanner {
		private BufferedReader br;
		private StringTokenizer st;

		public FastScanner() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		public String next() {
			while (st == null || !st.hasMoreTokens()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

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

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

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