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 *

class UnionRank:
  def __init__(self, n):
    self.n       = n
    self.parent  = range(n+1)
    self.rank    = [0]*(n+1)
    self.sizes   = [1]*(n+1)

  def union(self, x, y):
    xRoot = self.find(x)
    yRoot = self.find(y)
    if (xRoot == yRoot): return 0

    newRoot = xRoot if self.rank[xRoot] > self.rank[yRoot] else yRoot
    otherRoot = xRoot if newRoot==yRoot else yRoot
    
    result = self.sizes[newRoot]*self.sizes[otherRoot]
    self.parent[otherRoot] = newRoot
    self.sizes [newRoot] += self.sizes[otherRoot]
    if self.rank[xRoot] == self.rank[yRoot]:
      self.rank[newRoot] += 1

    return result

  def find(self, x):
    start = x
    path = []

    while self.parent[start] != start:
      path.append(start)
      start = self.parent[start]

    for elt in path:
      self.parent[elt] = start

    return start

  def same(self, x, y):
    return self.find(x)==self.find(y)

  def sets(self):
    sets = defaultdict(set)
    for i in xrange(1,self.n+1):
      sets[self.find(i)].add(i)
    return sets.values()

n,m = map(int, raw_input().split())
color = [UnionRank(n) for _ in xrange(101)]
for i in xrange(m):
  a,b,c = map(int, raw_input().split())
  color[c].union(a,b)

for i in xrange(int(raw_input())):
  a,b = map(int, raw_input().split())
  print sum(color[c].same(a,b) for c in xrange(101))
