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.*;



public class MrKitayutasColorGraph {

    static ArrayList<Edge> graph[];
    static boolean[] visited;
    static class Edge {
        int node;
        int weight;

        public Edge(int n, int w){
            node = n;
            weight = w;
        }

    }



    public static void main(String[] args){
        Scanner in = new Scanner(System.in);

        int nodes, edges;
        nodes = in.nextInt();
        edges = in.nextInt();

        graph = new ArrayList[nodes];
        visited = new boolean[nodes];

        for(int i = 0; i < nodes; ++i) graph[i] = new ArrayList<Edge>(); //Init Graph
        for(int i = 0; i < edges; ++i){
            int start, end, color;
            start = in.nextInt() - 1;
            end = in.nextInt() - 1;
            color = in.nextInt() - 1;

            graph[start].add(new Edge(end, color));
            graph[end].add(new Edge(start, color));
        }

        int queries = in.nextInt();
        for(int i = 0; i < queries; ++i){
            int start = in.nextInt() - 1;
            int end = in.nextInt() - 1;

            int cnt = 0;
            for(int k = 0; k < edges; ++k){
                clearVis();
                if(dfs(start, k, end)) cnt++;
            }
            System.out.println(cnt);
            clearVis();

        }

    }
    public static void clearVis(){
        for(int i =0; i < visited.length; ++i){
            visited[i] = false;
        }
    }

    public static boolean dfs(int node, int color , int dest){
        visited[node] = true;
        if(node == dest) return true;
        for(Edge tmp: graph[node]){
            if(tmp.weight == color && !visited[tmp.node]){
                if(dfs(tmp.node, color, dest)) return true;
            }
        }
        return false;
    }
}