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 main():
    n, m = map(int, input().split())
    edges = {_: {} for _ in range(1, m + 1)}
    for _ in range(m):
        a, b, c = map(int, input().split())
        d = edges[c]
        l = d.get(a)
        if l:
            l.append(b)
        else:
            d[a] = [b]
        l = d.get(b)
        if l:
            l.append(a)
        else:
            d[b] = [a]

    def dfs(t):
        unused[t] = False
        for y in dcolor.get(t, ()):
            if unused[y]:
                dfs(y)

    res = []
    n += 1
    unused = [True] * n
    for _ in range(int(input())):
        a, b = map(int, input().split())
        x = 0
        for dcolor in edges.values():
            for j in range(n):
                unused[j] = True
            dfs(a)
            if not unused[b]:
                x += 1
        res.append(str(x))
    print('\n'.join(res))


if __name__ == '__main__':
    main()
