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>.
from collections import defaultdict, deque

def bfs(g,s,c):
    visited = set()
    q = deque([s])
    while len(q) > 0:
        v = q.pop()
        visited.add(v)
        for adj,adj_c in g[v]:
            if adj_c == c and adj not in visited and adj not in q:
                q.append(adj)
    return visited
    
def con_components(g,c):
    visited = set()
    components = []
    for s in g.keys():
        if s not in visited:
            nodes = bfs(g,s,c)
            components.append(nodes)
            visited |= nodes
    return components

def main():
    n,m = map(int, input().split())
    g = defaultdict(list)
    colors = set()
    for _ in range(m):
        a,b,c = map(int, input().split())
        g[a].append((b,c))
        g[b].append((a,c))
        colors.add(c)
    components = {c:con_components(g,c) for c in colors}
    
    q = int(input())
    for _ in range(q):
        u,v = map(int, input().split())
        ct = 0
        for c in colors:
            for comp in components[c]:
                if u in comp and v in comp:
                    ct += 1
        print(ct)
    

if __name__ == '__main__':
    main()   