Mr. Kitayuta has just bought an undirected graph with n vertices and m edges. The vertices of the graph are numbered from 1 to n. Each edge, namely edge i, has a color ci, connecting vertex ai and bi.

Mr. Kitayuta wants you to process the following q queries.

In the i-th query, he gives you two integers - ui and vi.

Find the number of the colors that satisfy the following condition: the edges of that color connect vertex ui and vertex vi directly or indirectly.

Input

The first line of the input contains space-separated two integers - n and m(2 ≤ n ≤ 105, 1 ≤ m ≤ 105), denoting the number of the vertices and the number of the edges, respectively.

The next m lines contain space-separated three integers - ai, bi(1 ≤ ai < bi ≤ n) and ci(1 ≤ ci ≤ 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, (ai, bi, ci) ≠ (aj, bj, cj).

The next line contains a integer- q(1 ≤ q ≤ 105), denoting the number of the queries.

Then follows q lines, containing space-separated two integers - ui and vi(1 ≤ ui, vi ≤ n). It is guaranteed that ui ≠ vi.

Output

For each query, print the answer in a separate line.

Examples

Input

4 5
1 2 1
1 2 2
2 3 1
2 3 3
2 4 3
3
1 2
3 4
1 4


Output

2
1
0


Input

5 7
1 5 1
2 5 1
3 5 1
4 5 1
1 2 2
2 3 2
3 4 2
5
1 5
5 1
2 5
1 5
1 4


Output

1
1
1
1
2

Note

Let's consider the first sample. 

<image> The figure above shows the first sample. 

  * Vertex 1 and vertex 2 are connected by color 1 and 2. 
  * Vertex 3 and vertex 4 are connected by color 3. 
  * Vertex 1 and vertex 4 are not connected by any single color. 
def create_adj(edge):
    edges={}
    for u in edge:
        if u[0] in edges.keys():
            edges[u[0]].append(u[1])
        else:
            edges[u[0]]=[u[1]]
    return edges
def explore(adj,v,visited):
    visited[v]=True
    #print(visited)
   # print('adj',adj)
  #  print('v',v)
    for i in adj[v]:
        if visited[i]==False:
            explore(adj,i,visited)
def find(adj,u,v):
    global c
    if(u not in adj.keys() or v not in adj.keys()):
        return
    visited={}
    for r in adj.keys():
        visited[r]=False
    explore(adj,u,visited)
    if(visited[v]==True):

        c=c+1
 #       print('yes',c)
    return c
n,m=map(int,input().split())
ht={}
for i in range(m):
    a,b,c=map(int,input().split())
    if(c in ht.keys()):
        ht[c].append([a,b])
        ht[c].append([b,a])
    else:
        ht[c]=[[a,b],[b,a]]
q=int(input())
#print(ht)
c=0
for j in range(q):
    c=0
    u,v=map(int,input().split())
    for y in ht.keys():
        adj=create_adj(ht[y])

        #print(adj)
        find(adj,u,v)
    print(c)