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. 
from collections import defaultdict

def dfs(adjlist, start, visited, nodes, color):
    if start in visited:
        return
    
    visited.add(start)
    for (v,c) in adjlist[start]:
        if c == color:
            nodes[v-1] = nodes[start-1]
            dfs(adjlist, v, visited, nodes, color)
    
if __name__ =='__main__':
    n,m = map(int, input().split())
    adjlist = defaultdict(list)
    all_color = set()
    for i in range(0,m):
        tmp = list(map(int, input().split()))
        adjlist[tmp[0]].append((tmp[1],tmp[2]))
        adjlist[tmp[1]].append((tmp[0],tmp[2]))
        all_color.add(tmp[2])
    
    num_queries = int(input())
    queries = []
    for i in range(0,num_queries):
        queries.append(list(map(int, input().split())))
    
    color_map = {}
    for c in all_color:
        nodes = [i for i in range(1,n+1)]
        for src in range(1,n+1):
            visited = set()
            dfs(adjlist, src, visited, nodes, c)
            if len(visited) == n:
                break
        color_map[c] = nodes
        
    #print(color_map)
    for q in queries:
        count = 0
        for c in color_map.keys():
            if color_map[c][q[0]-1] == color_map[c][q[1]-1]:
                count += 1
        print(count)