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 build_graph():
    line1 = input().strip().split()
    n = int(line1[0])
    m = int(line1[1])
    graph = {}
    for _ in range(m):
        line = input().strip().split()
        u = int(line[0])
        v = int(line[1])
        c = int(line[2])
        if c not in graph:
            graph[c] = {j: [] for j in range(1, n+1)}
        graph[c][u].append(v)
        graph[c][v].append(u)
    return graph

parent_history = {}

def no_of_paths(u, v, graph):
    x = 0
    for c in graph:
        if c in parent_history:
            if v in parent_history[c]:
                parent = parent_history[c]
                if u in parent:
                    x += 1
            elif u in parent_history[c]:
                parent = parent_history[c]
                if v in parent:
                    x += 1
            else:
                parent = {}
                parent = dfs_visit(v, graph[c], parent)
                if len(parent_history[c]) < len(parent):
                    parent_history[c] = parent
                if u in parent:
                    x += 1
        else:
            parent = {}
            parent = dfs_visit(v, graph[c], parent)
            parent_history[c] = parent
            if u in parent:
                x += 1
    return x

def dfs_visit(i, adj_list, parent):
    for j in adj_list[i]:
        if j not in parent:
            parent[j] = i
            dfs_visit(j, adj_list, parent)
    return parent


if __name__ == "__main__":
    graph = build_graph()
    for _ in range(int(input())):
        line = input().strip().split()
        print(no_of_paths(int(line[0]), int(line[1]), graph))
