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>.



adj={}

n,m=list(map(int,input().split()))

for i in range(m):
  a,b,color=list(map(int,input().split()))
  if a not in adj:
    adj[a]=[]
  if b not in adj:
    adj[b]=[]
 
  adj[a].append((b,color))
  adj[b].append((a,color))

q_l=int(input())
q=[]
for i in range(q_l):

  q.append(list(map(int,input().split())))

visited=[0]*(n+1)


def dfs(u,v,color):
 
  global visited
  visited[u]=1
 
  if u==v:
    visited[u]=0
    return True 
  found=False
  for k,c in (adj[u]):
    
    if not visited[k] and c==color:
      
      s=dfs(k,v,color)
      
      if s:
        found=True
  visited[u]=0
  return found   




def main(q,n,m):
  
  global visited
  for i in range(len(q)):
    u,v=q[i]
    ans=0

    
    
    for j in range(1,m+1):
   
      
      found=dfs(u,v,j)
      if found:
        ans+=1
      
    print(ans)

    
      

    
main(q,n,m)