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.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Scanner;

/*

2 5
1 2 1
1 2 2
1 2 3
1 2 4
1 2 5
1
1 2

 */
public class b {

	static class Graph
	{
		class Edge
		{
			int v1;
			int v2;
			int color;
			
			Edge(int vv1, int vv2, int c)
			{
				v1 = vv1;
				v2 = vv2;
				color = c;
			}
			
			int other (int v)
			{
				if(v == v1)
					return v2;
				else if (v == v2)
					return v1;
				return 1/0;
			}
		}
		
		ArrayList<Edge>[] adjList;
		int size;
		int numColors;
		boolean[] colors;
		
		@SuppressWarnings("unchecked")
		Graph (int n, int m)
		{
			size = n;
			numColors = m;
			
			colors = new boolean[m];
			adjList = new ArrayList[n];
			
			for(int curr = 0; curr < n; ++curr)
			{
				adjList[curr] = new ArrayList<Edge>();
			}
		}
		
		void addEdge(int v1, int v2, int c)
		{
			Edge e = new Edge(v1, v2, c);
			adjList[v1].add(e);
			adjList[v2].add(e);
			
			colors[c] = true;
		}
		
		int count(int v1, int v2)
		{
			int count = 0;
			
			for(int currC = 0; currC < numColors; ++currC)
			{
				if(colors[currC] && bfs(v1, v2, currC))
				{
					++count;
				}
			}
			
			return count;
		}
		
		boolean bfs(int start, int end, int color)
		{
			boolean[] visited = new boolean[size];
			
			ArrayDeque<Integer> q = new ArrayDeque<Integer>();
			q.add(start);
			visited[start] = true;
			
			while(!q.isEmpty())
			{
				int node = q.poll();
				
				if(node == end)
					return true;
				
				for(Edge e: adjList[node])
				{
					if(e.color != color)
						continue;
					
					int next = e.other(node);
					if(!visited[next])
					{
						visited[next] = true;
						q.add(next);
					}
				}
			}
			
			return false;
		}
	}
	
	public static void main(String[] args) 
	{
		Scanner in = new Scanner(System.in);
		
		int n = in.nextInt();
		int m = in.nextInt();
		
		Graph g = new Graph(n, m);
		
		for(int currEdge = 0; currEdge < m; ++currEdge)
		{
			g.addEdge(in.nextInt()-1, in.nextInt()-1, in.nextInt()-1);
		}
		
		int q = in.nextInt();
		
		for(int currQ = 0; currQ < q; ++currQ)
		{
			System.out.println(g.count(in.nextInt()-1, in.nextInt()-1));
		}
		
		in.close();
	}

}
