Mr. Kitayuta has just bought an undirected graph consisting of *n* vertices and *m* edges. The vertices of the graph are numbered from 1 to *n*. Each edge, namely edge *i*, has a color *c*<sub class="lower-index">*i*</sub>, connecting vertex *a*<sub class="lower-index">*i*</sub> and *b*<sub class="lower-index">*i*</sub>.

Mr. Kitayuta wants you to process the following *q* queries.

In the *i*-th query, he gives you two integers — *u*<sub class="lower-index">*i*</sub> and *v*<sub class="lower-index">*i*</sub>.

Find the number of the colors that satisfy the following condition: the edges of that color connect vertex *u*<sub class="lower-index">*i*</sub> and vertex *v*<sub class="lower-index">*i*</sub> directly or indirectly.
The first line of the input contains space-separated two integers — *n* and *m* (2 ≤ *n* ≤ 100, 1 ≤ *m* ≤ 100), denoting the number of the vertices and the number of the edges, respectively.

The next *m* lines contain space-separated three integers — *a*<sub class="lower-index">*i*</sub>, *b*<sub class="lower-index">*i*</sub> (1 ≤ *a*<sub class="lower-index">*i*</sub> &lt; *b*<sub class="lower-index">*i*</sub> ≤ *n*) and *c*<sub class="lower-index">*i*</sub> (1 ≤ *c*<sub class="lower-index">*i*</sub> ≤ *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*, (*a*<sub class="lower-index">*i*</sub>, *b*<sub class="lower-index">*i*</sub>, *c*<sub class="lower-index">*i*</sub>) ≠ (*a*<sub class="lower-index">*j*</sub>, *b*<sub class="lower-index">*j*</sub>, *c*<sub class="lower-index">*j*</sub>).

The next line contains a integer — *q* (1 ≤ *q* ≤ 100), denoting the number of the queries.

Then follows *q* lines, containing space-separated two integers — *u*<sub class="lower-index">*i*</sub> and *v*<sub class="lower-index">*i*</sub> (1 ≤ *u*<sub class="lower-index">*i*</sub>, *v*<sub class="lower-index">*i*</sub> ≤ *n*). It is guaranteed that *u*<sub class="lower-index">*i*</sub> ≠ *v*<sub class="lower-index">*i*</sub>.
import sys,os,io
import math,bisect,operator
inf,mod = float('inf'),10**9+7
# sys.setrecursionlimit(10 ** 6)
from itertools import groupby,accumulate
from heapq import heapify,heappop,heappush
from collections import deque,Counter,defaultdict
I = lambda : int(sys.stdin.buffer.readline())
Neo = lambda : list(map(int, sys.stdin.buffer.readline().split()))

n,m = Neo()
Color = defaultdict(set)

G = defaultdict(set)    

C = set()

def addEdge(a,b):
    G[a].add(b)
    G[b].add(a)
    
def addColor(a,b,c):
    Color[(a,b)].add(c)
    Color[(b,a)].add(c)
    C.add(c)
    
for i in range(m):
    a,b,c = Neo()
    addColor(a,b,c)
    addEdge(a,b)
    
vis = [False]*(n+1)
Ans = []
    
def fun(node,dest,vis,grp):
    if not grp:
        return
    if node == dest:
        for i in grp:
            Ans.append(i)
        return
    vis[node] = True
    for i in G.get(node,[]):
        if not vis[i]:
            newvis = vis.copy()
            z = grp.intersection(Color[node,i])
            fun(i,dest,newvis,z)

for i in range(I()):
    a,b = Neo()
    vis = [False]*(n+1)
    grp = C.copy()
    fun(a,b,vis,grp)
    print(len(set(Ans)))
    Ans =[]

    