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.List;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.HashMap;
import java.io.BufferedReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.util.LinkedList;
import java.util.HashSet;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 * @author Kartik Singal (ka4tik)
 */
public class Main {
	public static void main(String[] args) {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		InputReader in = new InputReader(inputStream);
		PrintWriter out = new PrintWriter(outputStream);
		TaskD solver = new TaskD();
		solver.solve(1, in, out);
		out.close();
	}
}

class TaskD {

    public long hashCode(int a,int b)
    {
        return (long)a*10000000+(long)b;
    }
    public void solve(int testNumber, InputReader in, PrintWriter out) {
        int n=in.nextInt();
        int m=in.nextInt();

        HashMap<Long,Integer> mapper=new HashMap<Long, Integer>();
        HashSet<Integer> colors[]=new HashSet[n];
        for(int i=0;i<colors.length;i++)
            colors[i]=new HashSet<Integer>();
        List<Integer> us=new LinkedList<Integer>();
        List<Integer> vs=new LinkedList<Integer>();

        int counter=0;
        for(int i=0;i<m;i++)
        {
            int u=in.nextInt();u--;
            int v=in.nextInt();v--;
            int c=in.nextInt();c--;
            colors[u].add(c);
            colors[v].add(c);
            if(!mapper.containsKey(hashCode(u,c)))
                mapper.put(hashCode(u,c),counter++);
            if(!mapper.containsKey(hashCode(v,c)))
                mapper.put(hashCode(v,c),counter++);

            u=mapper.get(hashCode(u,c));
            v=mapper.get(hashCode(v,c));
            us.add(u);
            vs.add(v);
        }
        UnionFind uf=new UnionFind(counter);
        for(int i=0;i<vs.size();i++)
            uf.union(us.get(i), vs.get(i));
        int q=in.nextInt();
        HashMap<Long,Integer> answers=new HashMap<Long, Integer>();
        while(q-->0)
        {
            int u=in.nextInt();u--;
            int v=in.nextInt();v--;
            if(u>v)
            {
                int temp=u;u=v;v=temp;
            }
            if(colors[v].size()<colors[u].size())
            {
                int temp=u;u=v;v=temp;
            }
            if(answers.containsKey(hashCode(u,v))) {
                out.println(answers.get(hashCode(u,v)));
                continue;
            }
            int ans=0;
            for(Integer c:colors[u])
            {
                Integer a=mapper.get(hashCode(u,c));
                Integer b=mapper.get(hashCode(v,c));
                if(a==null||b==null)
                    continue;
                if(uf.connected(a,b))
                    ans++;
            }
            out.println(ans);
            answers.put(hashCode(u,v),ans);
        }
    }
}

class InputReader {
    public BufferedReader reader;
    public StringTokenizer tokenizer;

    public InputReader(InputStream stream) {
        reader = new BufferedReader(new InputStreamReader(stream), 32768);
        tokenizer = null;
    }

    public String next() {
        while (tokenizer == null || !tokenizer.hasMoreTokens()) try {
            tokenizer = new StringTokenizer(reader.readLine());
        } catch (IOException e) {
            throw new RuntimeException(e);
        }
        return tokenizer.nextToken();
    }

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

}

class UnionFind {
    private int[] id;     // id[i] = parent of i
    private byte[] rank;  // rank[i] = rank of subtree rooted at i (cannot be more than 31)
    private int count;    // number of components

    public UnionFind(int N) {
        if (N < 0) throw new IllegalArgumentException();
        count = N;
        id = new int[N];
        rank = new byte[N];
        for (int i = 0; i < N; i++) {
            id[i] = i;
            rank[i] = 0;
        }
    }

    public int find(int p) {
        if (p < 0 || p >= id.length) throw new IndexOutOfBoundsException();
        while (p != id[p]) {
            id[p] = id[id[p]];    // path compression by halving
            p = id[p];
        }
        return p;
    }

    public boolean connected(int p, int q) {
        return find(p) == find(q);
    }

    public void union(int p, int q) {
        int i = find(p);
        int j = find(q);
        if (i == j) return;

        // make root of smaller rank point to root of larger rank
        if (rank[i] < rank[j]) id[i] = j;
        else if (rank[i] > rank[j]) id[j] = i;
        else {
            id[j] = i;
            rank[i]++;
        }
        count--;
    }

}

