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.*;
import java.util.*;
public class Main{
	private static BufferedReader br  = new BufferedReader(new InputStreamReader(System.in));
	private static PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
	public static void main(String[] args) throws IOException {
		solve();
		out.close();
	}
	static class Node{
		int v, t;
		Node(int v, int t){
			this.v = v;
			this.t = t;
		}
	}
	static int MAX = Integer.MAX_VALUE;
	static void solve() throws IOException {
		String str = br.readLine();
		StringTokenizer st = new StringTokenizer(str);
		int n = Integer.parseInt(st.nextToken());	
		int m = Integer.parseInt(st.nextToken());	
		ArrayList<Node>[] adj = new ArrayList[n + 1];
		for (int i = 0; i <= n; i++) {
			adj[i] = new ArrayList<Node>();
		}
		for (int i = 0; i < m; i++) {
			str = br.readLine();
			st = new StringTokenizer(str);
			int u = Integer.parseInt(st.nextToken());
			int v = Integer.parseInt(st.nextToken());
			int p = Integer.parseInt(st.nextToken());
			adj[u].add(new Node(v, p));
			adj[v].add(new Node(u, p));
		}
		int k = Integer.parseInt(br.readLine());
		boolean[] visited = new boolean[n+1];
		boolean[] color = new boolean[m+1];
		int count = 0;
		for (int i = 0; i < k; i++) {
			st = new StringTokenizer(br.readLine());
			int s = Integer.parseInt(st.nextToken());
			int d = Integer.parseInt(st.nextToken());
			Arrays.fill(visited, false);
			Arrays.fill(color, false);
			count = 0;
			visited[s] = true;
			count = dfs(s, adj, visited, d, -1, count, color);
			out.println("" + count);
		}
	}
	static int dfs(int u, ArrayList<Node>[] adj,boolean[] visited, int d, int c, int count, boolean[] color){
		for (Node n : adj[u]) {
			int v = n.v;
			if (visited[v] == false && (n.t == c || c == -1) ){
				// out.print(v + " ");
				if (v == d && color[n.t] == false) {
					color[n.t] = true;
					count++;
				}
				else{
					visited[v] = true;
					count = dfs(v, adj, visited, d, n.t, count, color);
					visited[v] = false;
				}
			}
		}
		return count;
	}
}