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>.
class Solution:
    def __init__(self, n, m, edges):
        self.n = n
        self.m = m
        self.edges = [[[] for _ in range(m+1)] for _ in range(n+1)]
        for _from, _to, _color in edges:
            self.edges[_from][_color].append(_to)
            self.edges[_to][_color].append(_from)
        self.id = 0

    def initialize(self):
        self.discovered = [[0]*(self.n+1) for i in range(self.m+1)]

    def bfs(self, x0, color):
        self.id += 1
        stack = [x0]
        while stack:
            n = stack.pop()
            if not self.discovered[color][n]:
                self.discovered[color][n] = self.id
                for neigh in self.edges[n][color]:
                    stack.append(neigh)

def solve(graph, colors, u, v):
    counter = 0
    for color in colors:
        if not graph.discovered[color][u] and not graph.discovered[color][v]:
            graph.bfs(u, color)
        if graph.discovered[color][u] == graph.discovered[color][v]:
            counter += 1
    return counter

def main():
    n, m = map(int, input().split())
    edges = []
    colors = set()
    for _ in range(m):
        a, b, c = map(int, input().split())
        colors.add(c)
        edges.append((a, b, c))
    graph = Solution(n, m, edges)
    graph.initialize()
    q = int(input())
    for _ in range(q):
        u, v = map(int, input().split())
        print(solve(graph, colors, u, v))

if __name__ == '__main__':
    main()
        

    
        
