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>.
#code by aanchaltiwari__
def dfs(x, y, color):
    visited = [False for i in range(n + 1)]
    visited[x] = True
    stack = [x]
    while stack:
        node = stack.pop()
        for child, childcolor in g[node]:
            if visited[child] == False and childcolor == color:
                visited[child] = True
                stack.append(child)
    if visited[y] == True:
        return True
    return False


n, m = map(int, input().split())
g = {i: [] for i in range(1, n + 1)}
s = set()
for i in range(m):
    u, v, c = map(int, input().split())
    s.add(c)
    g[u].append([v, c])
    g[v].append([u, c])
s = list(s)
# print(g)
# print(s)
ans = [[0 for i in range(n+1)]for j in range(n+1)]
for i in range(1, n + 1):
    for j in range(1, n + 1):
        for c in s:
            if i != j:
                if dfs(i, j, c):
                    ans[i][j] += 1
# print(ans)
q = int(input())
for i in range(q):
    u, v = map(int, input().split())
    print(ans[u][v])
