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.*;

import static java.lang.Math.*;

public class Main extends PrintWriter {
    BufferedReader in;
    StringTokenizer stok;
    final Random rand = new Random(31);
    final int inf = (int) 1e9;
    final long linf = (long) 1e18;

    final static String IO = "_std";

    class Edge {
        int to, c;

        public Edge(int to, int c) {
            this.to = to;
            this.c = c;
        }
    }

    List<Edge>[] g;
    int[][] used;
    int time = 0;

    int dfs(int v, int f, int c) {
        used[v][c] = time;
        if (v == f) {
            return 1;
        }
        int res = 0;
        for (Edge e : g[v]) {
            if (used[e.to][e.c] != time && (c == 0 || e.c == c)) {
                res += dfs(e.to, f, e.c);
            }
        }
        return res;
    }

    void solve() throws IOException {
        int n = nextInt();
        int m = nextInt();
        g = createAdjacencyList(n);
        for (int i = 0; i < m; i++) {
            int x = nextInt() - 1;
            int y = nextInt() - 1;
            int c = nextInt();
            g[x].add(new Edge(y, c));
            g[y].add(new Edge(x, c));
        }
        used = new int[n][m + 1];
        int q = nextInt();
        for (; q-- > 0; ) {
            int x = nextInt() - 1;
            int y = nextInt() - 1;
            time++;
            println(dfs(x, y, 0));
        }
    }

    Main() {
        super(System.out);
        in = new BufferedReader(new InputStreamReader(System.in));
    }

    Main(String fileIn, String fileOut) throws IOException {
        super(fileOut);
        in = new BufferedReader(new FileReader(fileIn));
    }

    public static void main(String[] args) throws IOException {
        Main main;
        if ("_std".equals(IO)) {
            main = new Main();
        } else if ("_iotxt".equals(IO)) {
            main = new Main("input.txt", "output.txt");
        } else {
            main = new Main(IO + ".in", IO + ".out");
        }
        main.solve();
        main.close();
    }

    String next() throws IOException {
        while (stok == null || !stok.hasMoreTokens()) {
            String s = in.readLine();
            if (s == null) {
                return null;
            }
            stok = new StringTokenizer(s);
        }
        return stok.nextToken();
    }

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

    long nextLong() throws IOException {
        return Long.parseLong(next());
    }

    double nextDouble() throws IOException {
        return Double.parseDouble(next());
    }

    int[] nextIntArray(int len) throws IOException {
        int[] a = new int[len];
        for (int i = 0; i < len; i++) {
            a[i] = nextInt();
        }
        return a;
    }

    int[] nextIntArraySorted(int len) throws IOException {
        int[] a = nextIntArray(len);
        shuffle(a);
        Arrays.sort(a);
        return a;
    }

    void shuffle(int[] a) {
        for (int i = 1; i < a.length; i++) {
            int x = rand.nextInt(i + 1);
            int _ = a[i];
            a[i] = a[x];
            a[x] = _;
        }
    }

    void shuffleAndSort(int[] a) {
        shuffle(a);
        Arrays.sort(a);
    }

    boolean nextPermutation(int[] p) {
        for (int a = p.length - 2; a >= 0; --a) {
            if (p[a] < p[a + 1])
                for (int b = p.length - 1; ; --b)
                    if (p[b] > p[a]) {
                        int t = p[a];
                        p[a] = p[b];
                        p[b] = t;
                        for (++a, b = p.length - 1; a < b; ++a, --b) {
                            t = p[a];
                            p[a] = p[b];
                            p[b] = t;
                        }
                        return true;
                    }
        }
        return false;
    }

    <T> List<T>[] createAdjacencyList(int n) {
        List<T>[] res = new List[n];
        for (int i = 0; i < n; i++) {
            res[i] = new ArrayList<>();
        }
        return res;
    }
}