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 math,sys,bisect,heapq
from collections import defaultdict,Counter,deque
from itertools import groupby,accumulate
#sys.setrecursionlimit(200000000)
int1 = lambda x: int(x) - 1
#def input(): return sys.stdin.readline().strip()
input = iter(sys.stdin.buffer.read().decode().splitlines()).__next__
ilele = lambda: map(int,input().split())
alele = lambda: list(map(int, input().split()))
ilelec = lambda: map(int1,input().split())
alelec = lambda: list(map(int1, input().split()))
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
#MOD = 1000000000 + 7
def Y(c):  print(["NO","YES"][c])
def y(c):  print(["no","yes"][c])
def Yy(c):  print(["No","Yes"][c])

n,m = ilele()
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 = ilele()
    addColor(a,b,c)
    addEdge(a,b)
    
vis = [False]*(n+1)
Ans = []
    
def fun(node,dest,vis,grp):
    #print(node,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,[]):
        #print(i)
        if not vis[i]:
            newvis = vis.copy()
            #newvis[c] =True
            z = grp.intersection(Color[node,i])
            #print(z)
            fun(i,dest,newvis,z)
            
#print(Color,G)
for i in range(int(input())):
    a,b = ilele()
    vis = [False]*(n+1)
    grp = C.copy()
    fun(a,b,vis,grp)
    #print(Ans)
    print(len(Ans))
    Ans =[]
 