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. 

visited = {}

def dfs(L,a,b):
    if (a == b):
        return 1
    visited[a] = 1
    c = False
    for i in L[a]:
        if (visited.has_key(i)):
            continue
        else:
            c = c or dfs(L,i,b)
    return c



def main():
    global visited
    n, m = [int(i) for i in raw_input().split()]
    L = [[[] for j in range(n)] for i in range(m)]
    for i in range(m):
        a, b, c = [int(i) for i in raw_input().split()]
        a -= 1
        b -= 1
        c -= 1
        L[c][a].append(b)
        L[c][b].append(a)
    q = input()
    for i in range(q):
        count = 0
        a, b = [int(i) for i in raw_input().split()]
        a -= 1
        b -= 1
        for j in range(m):
            visited = {}
            if dfs(L[j], a, b):
                count += 1
        print count




if __name__ == '__main__':
    main()