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.PrintWriter;
import java.util.ArrayList;
import java.util.StringTokenizer;
import java.util.TreeSet;

public class MrKitayutasColorfulGraph {
	static final StdIn in = new StdIn();
	static final PrintWriter out = new PrintWriter(System.out);

	public static void main(String[] args) {
		new Solver();
		out.close();
	}

	@SuppressWarnings("unchecked")
	private static class Solver {
		int n, m;
		ArrayList<Integer>[][] adj;
		TreeSet<Integer> uniqueColors;

		public Solver() {
			n = in.nextInt();
			m = in.nextInt();
			adj = new ArrayList[n + 1][n + 1];
			for (int i = 0; i <= n; ++i) {
				for (int j = 0; j <= n; ++j)
					adj[i][j] = new ArrayList<>();
			}
			for (int i = 0; i < m; ++i) {
				int u = in.nextInt(), v = in.nextInt(), color = in.nextInt();
				adj[u][v].add(color);
				adj[v][u].add(color);
			}
			int q = in.nextInt();
			for (int i = 0; i < q; ++i) {
				boolean[][] visited = new boolean[n + 1][n + 1];
				uniqueColors = new TreeSet<>();
				int u = in.nextInt(), v = in.nextInt();
				visited[u][u] = true;
				dfs(u, v, -1, visited);
				out.println(uniqueColors.size());
			}
		}

		boolean dfs(int from, int to, int curColor, boolean[][] visited) {
			if (from == to) {
				return true;
			}
			for (int i = 1; i <= n; ++i) {
				if (visited[from][i])
					continue;
				visited[from][i] = visited[i][from] = true;
				for (int j = 0; j < adj[from][i].size(); ++j) {
					int c = adj[from][i].get(j);
					if (curColor == -1) {
						boolean found = dfs(i, to, c, visited);
						if (found)
							uniqueColors.add(c);
					} else if (curColor == c) {
						boolean found = dfs(i, to, curColor, visited);
						if (found)
							uniqueColors.add(c);
					}
				}
				visited[from][i] = visited[i][from] = false;
			}
			return false;
		}
	}

	private static class StdIn {
		private BufferedReader br;
		private StringTokenizer st;

		public StdIn() {
			st = null;
			br = new BufferedReader(new InputStreamReader(System.in), 1 << 20);
		}

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

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