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 sets import Set

visited = []
graph = {}
cores = Set()

def has_c_path(u, v, c):
  visited[u] = True
  if u == v: return 1
  for x in graph[u]:
    if x[1] != c: continue
    if visited[x[0]] == True: continue
    resp = has_c_path(x[0], v, c)
    if resp == 1 : return 1
    
  return 0 

s = raw_input().split(" ")
n = int(s[0])
m = int(s[1])

for i in range(1, n+1):
  graph[i] = []

for i in range(m):
  s = raw_input().split(" ")
  a = int(s[0])
  b = int(s[1])
  c = int(s[2])
  cores.add(c)
  graph[b].append((a, c))
  graph[a].append((b, c))

q = int(raw_input())
for i in range(q):
  ans = 0
  s = raw_input().split(" ")
  u = int(s[0])
  v = int(s[1])
  for cor in cores:
    visited = [False] * (n + 1)
    result = has_c_path(u, v, cor)
    ans += result
  print ans
  
