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 sys import stdin

import fileinput
import string
import sys
from collections import defaultdict

inp1 = map(int, stdin.readline().rstrip().split())
count1 = inp1[1]
graph = []
g = defaultdict(lambda: ([], [])) # dest, col
allcols = set()
for i in xrange(count1):
	line = map(int, stdin.readline().rstrip().split())
	graph.append(line)
	t = g[line[0]]
	t[0].append(line[1])
	t[1].append(line[2])
	t = g[line[1]]
	t[0].append(line[0])
	t[1].append(line[2])
	allcols.add(line[2])

count2 = int(stdin.readline().rstrip())
queries = []


def bfsAAA(start, dest, col):
	if start == dest:
		return 0
	seen = set([start])
	best = None
	cur = g[start]
	newcur = ([],[])
	for i in xrange(len(cur[0])):
		node = cur[0][i]
		if node == dest and col == cur[1][i]:
			return 1
		if node != dest and col == cur[1][i]:
			seen.add(node)
			newcur[0].append(node)
			newcur[1].append(cur[1][i])
	cur = newcur
	traversed = 1
	#print cur
	while True:
		newcur = ([],[])
		for i in xrange(len(cur[0])):
			node = cur[0][i]
			neighbors = g[node]
			neighnodes = neighbors[0]
			neighcols = neighbors[1]
			for j in xrange(len(neighnodes)):
				if neighnodes[j] == dest and neighcols[j] == col:
					return traversed + 1
				if neighnodes[j] not in seen and neighcols[j] == col and neighcols[j] != dest:
					seen.add(neighnodes[j])
					newcur[0].append(neighnodes[j])
					newcur[1].append(neighcols[j])
		if len(newcur[0]) == 0:
			break
		cur = newcur
		traversed += 1
	return None


def bfs(start, dest, col, seen=set(), tot=0):
	newnodes = []
	for i in start:
		tot += 1
		cur = g[i]
		neighnodes = cur[0]
		neighcols = cur[1]
		#print neighnodes, neighcols
		for idx, n in enumerate(neighnodes):
			#print seen#n, dest, neighcols[idx], col, n in seen
			if n == dest and neighcols[idx] == col:
				return tot
			if n not in seen and neighcols[idx] == col:
				newnodes.append(n)
				seen.add(n)
	if len(newnodes) == 0:
		return None
	return bfs(newnodes, dest, col, seen, tot + 1)



for i in xrange(count2):
	line = map(int, stdin.readline().rstrip().split())
	tot = 0
	for color in allcols:
		#print [line[0]], line[1], color
		r = bfs([line[0]], line[1], color, set())
		if r != None:
			tot += 1
	print tot




