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. 
# 505B
from collections import defaultdict

__author__ = 'artyom'

read = lambda: map(int, input().split())
all_edges = defaultdict(list)
n, m = read()
for _ in range(m):
  a, b, c = read()
  all_edges[c].append((a, b))

graphs = []


def root(components, x, l=0):
  return (x, l) if components[x] == x else root(components, components[x], l + 1)


def connected_components(edges):
  components = [i for i in range(n + 1)]
  for u, v in edges:
    ru, lu = root(components, u)
    rv, lv = root(components, v)
    if ru != rv:
      if lu < lv:
        components[ru] = rv
      else:
        components[rv] = ru
  return components


for c, edges in all_edges.items():
  graphs.append(connected_components(edges))


def count_in_same_component(u, v):
  count = 0
  for graph in graphs:
    if root(graph, u)[0] == root(graph, v)[0]:
      count += 1
  return count


ans = ''
for __ in range(int(input())):
  u, v = read()
  ans += str(count_in_same_component(u, v)) + '\n'
print(ans)