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. 
import sys
from collections import defaultdict
from functools import lru_cache
from collections import Counter

def mi(s):
    return map(int, s.strip().split())

def lmi(s):
    return list(mi(s))

def mf(f, s):
    return map(f, s)

def lmf(f, s):
    return list(mf(f, s))

def path(graph, u, v, color, visited):
    if u == v:
        return True
    elif u in visited:
        return False

    visited.add(u)
    for child, c in graph[u]:
        if color == c and path(graph, child, v, color, visited):
            return True
    return False

def main(graph, queries, colors):
    ans = []
    for u, v in queries:
        c = 0
        for color in colors:
            if path(graph, u, v, color, set()):
                c += 1
        ans.append(c)

    for n in ans:
        print(n)

if __name__ == "__main__":
    queries = []
    colors = set()
    for e, line in enumerate(sys.stdin.readlines()):
        if e == 0:
            n, ed = mi(line)
            graph = defaultdict(list)
            for i in range(1, n + 1):
                graph[i]
            k = 0
        elif ed > 0:
            a, b, c = mi(line)

            # Undirected graph.
            graph[a].append((b, c))
            graph[b].append((a, c))
            colors.add(c)
            ed -= 1
        elif ed == 0:
            ed -= 1
            continue
        else:
            queries.append(lmi(line))
    main(graph, queries, colors)
