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. 
def dfs(Map,u,v,color):
    count=0
    for c in color:
        visited=[0 for i in range(110)]
        l=[]
        l.append(u)
        visited[u]=1
        while len(l)!=0:
            temp=l[len(l)-1]
            flag=1
            for i in range(len(Map[0])):
                if c in Map[temp][i] and visited[i]==0:
                    visited[i]=1
                    l.append(i)
                    flag=0
            if flag:
                l.pop()
        if visited[v]:
            count+=1

    return count

if __name__ == '__main__':
    R=lambda:map(int,raw_input().split())
    n,m=R()
    Map=[[[] for i in range(110)] for i in range(110)]
    color=[]
    for i in range(m):
        a,b,c=R()
        Map[a][b].append(c)
        Map[b][a].append(c)
        if c not in color:
            color.append(c)
    q=int(raw_input())
    ans=[]
    for i in range(q):
        u,v=R()
        ans.append(dfs(Map,u,v,color))

    for i in ans:
        print i
