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>.
def dfs(v, color, finish):
    global helper
    if v == finish:
        helper += 1
        return 
    used[v] = True
    for i in m[v]:
        if not used[i[0]] and i[1] == color:
           dfs(i[0], color, finish)


f = open('input.txt')
n, k = map(int, input().split())
m = [[] for i in range(n)]
for i in range(k):
    x, y, z = map(int, input().split())
    m[x - 1].append((y - 1, z))
    m[y - 1].append((x - 1, z))
k = int(input())
for i in range(k):
    x, y = map(int, input().split())
    helper = 0
    sett = set()
    for j in m[x - 1]:
        used = [False] * n
        if j[1] not in sett:
            sett.add(j[1])
            dfs(x - 1, j[1], y - 1)
    print(helper)


                
