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


def make(n):
  m = {}
  for i in range(n):
    m[i] = i
  return m

def union(m, a, b):
  pa = find(m, a)
  pb = find(m, b)
  m[pa] = pb

def find(m, a):
  if m[a] == a:
    return a
  else:
    ans = find(m, m[a])
    m[a] = ans
    return ans

N, M = map(int, sys.stdin.readline().split())
ms = []
for _ in range(M):
  m = make(N)
  ms.append(m)

for _ in range(M):
  sys.stdin.readline
  a, b, c = map(lambda f: int(f)-1, sys.stdin.readline().split())
  union(ms[c], a, b)

Q = int(sys.stdin.readline())
for _ in range(Q):
  u, v = map(lambda f: int(f)-1, sys.stdin.readline().split())
  ans = 0
  for m in ms:
    if find(m, u) == find(m, v):
      ans += 1
  print ans

