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. 
# http://codeforces.com/contest/505/problem/B
class Node:
	def __init__(self, n):
		self.n = n
		self.neighs = {}

	def dfs_color(self, f, t, nodes, visited):
		# if self.n in visited:
			# return visited[self.n]
		visited.add(self.n)
		connected = set()

		for neigh, colors_to_neigh in self.neighs.items():
			if neigh == f or neigh in visited:
				continue
			colors = colors_to_neigh if neigh == t else nodes[neigh].dfs_color(self.n,t,nodes, visited)
			connected.update(colors.intersection(colors_to_neigh))
		# visited[self.n] = connected
		visited.remove(self.n)
		return connected

	def dfs_colored_path(self, cc, nodes, visited):
		visited.add(self.n)
		for neigh, colors_to_neigh in self.neighs.items():
			if neigh in visited or cc not in colors_to_neigh:
				continue
			nodes[neigh].dfs_colored_path(cc, nodes, visited)
		return visited

	def search_with_color(self, t, nodes):
		colors_to_explore = set()
		colors = set()

		for neigh, colors_to_neigh in self.neighs.items():
			colors_to_explore.update(colors_to_neigh)

		for color in colors_to_explore:
			cities = self.dfs_colored_path(color, nodes, set())
			if t in cities:
				colors.add(color)
		return len(colors)



n, m = [int(n) for n in input().split()]
# edges = [int(n) for _ in range(m) for input().split()]
edges = []
while m:
	edge = [int(n) for n in input().split()]
	edges.append(edge)
	m-=1


nodes = [Node(i) for i in range(n+1)]
for f, t, c in edges:
	if t not in nodes[f].neighs:
		nodes[f].neighs[t] = set() 
	nodes[f].neighs[t].add(c)

	f,t = t,f
	if t not in nodes[f].neighs:
		nodes[f].neighs[t] = set() 
	nodes[f].neighs[t].add(c)

qs = int(input())

for q in range(qs):
	# visited = set()
	f, t = [int(n) for n in input().split()]
	# print(len(nodes[f].dfs_color(f, t, nodes, visited)))
	print(nodes[f].search_with_color(t, nodes))
