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. 
def solution():
	n,m = [int(x) for x in input().split(' ')]
	graphs = {}
	edges = {}
	for i in range(m):
		x,y,c = input().split(' ')
		if c not in graphs:
			graphs[c]={}
		if x not in graphs[c]:
			graphs[c][x] = []

		if y not in graphs[c]:
			graphs[c][y] = []

		graphs[c][x].append(y)
		graphs[c][y].append(x)

	q = int(input())
	queries = []
	for i in range(q):
		x,y = input().split(' ')
		ans = 0
		for c,graph in graphs.items():
			ans+=1 if areConnected(x,y,graph) else 0

		print(ans)

def areConnected(x,y,graph):
	if x not in graph or y not in graph:
		return False
	queu = [x]
	already = [x]
	while len(queu) != 0:
		current = queu[0]
		if current == y:
			return True
		del queu[0]
		already.append(current)
		for i in graph[current]:
			if i not in already and i not in queu:
				if i == y:
					return True

				queu.append(i)

	return False

solution()