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. 
n, m = [int(i) for i in raw_input().split()]
graph = {}
colors = set()
for i in range(1, n + 1):
	graph[i] = {}
for i in range(m):
	a, b, c = [int(i) for i in raw_input().split()]
	colors.add(c)
	if c in graph[a]:
		graph[a][c].append(b)
	else:
		graph[a][c] = [b]
	if c in graph[b]:
		graph[b][c].append(a)
	else:
		graph[b][c] = [a]
q = int(raw_input())

def dfs(u, v, c):
	visited, stack = set(), [u]
	while stack:
		node = stack.pop()
		if node not in visited:
			visited.add(node)
			if c in graph[node]:
				for neighbor in graph[node][c]:
					if neighbor == v:
						return True
					stack.append(neighbor)
	return False
	
for i in range(q):
	u, v = [int(i) for i in raw_input().split()]
	count = 0
	for c in colors:
		if dfs(u, v, c):
			count += 1
	print count
