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 TestClass {
    private static final boolean[] used = new boolean[101];

    public static void main(String[] args) {
        int counter = 0;
        Map<Integer, Integer> colorToIndex = new HashMap<>();
        List<Graph> graphs = new ArrayList<>();
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        int m = scanner.nextInt();
        for (int i = 0 ; i < m ; ++i) {
            int from = scanner.nextInt() - 1;
            int to = scanner.nextInt() - 1;
            int color = scanner.nextInt();
            if (!colorToIndex.containsKey(color)) {
                Graph graph = new Graph(n);
                graphs.add(graph);
                colorToIndex.put(color, counter++);
            }

            int graphId = colorToIndex.get(color);
            graphs.get(graphId).addEdge(from, to);
            graphs.get(graphId).addEdge(to, from);

        }


        StringBuilder stringBuilder = new StringBuilder();
        int q = scanner.nextInt();
        for (int i = 0 ; i < q; ++i) {
            int answer = 0;
            int from = scanner.nextInt() - 1;
            int to = scanner.nextInt() - 1;
            for (Graph graph : graphs) if (graph.checkDest(from, to)) answer++;
            stringBuilder.append(answer).append('\n');
        }
        System.out.println(stringBuilder);
    }

    private static final class Graph{
        private final List<List<Integer>> g;
        private Graph(int n) {
            g = new ArrayList<>();
            for (int i = 0 ; i < n ; ++i) g.add(new ArrayList<>());
        }

        private void addEdge(int from , int to) {
            get(from).add(to);
        }

        private List<Integer> get(int i) {return g.get(i);}

        private boolean checkDest(int from , int to) {
            Arrays.fill(used, false);
            return dfs(from, to);
        }

        private boolean dfs(int vert, int to) {
            if(vert == to) return true;
            used[vert] = true;
            for (int v : get(vert)) if (!used[v]) if (dfs(v, to)) return true;
            return false;
        }

    }

}
