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. 
import copy
info= list(map(int, input().split()))
n=info[0]
m=info[1]
mat=[[[] for i in range(n)] for j in range(n)]
for i in range(m):
    info= list(map(int, input().split()))
    a=info[0]
    b=info[1]
    c=info[2]
    mat[a-1][b-1].append(c)
    mat[b-1][a-1].append(c)
q=int(input())
for i in range(q):
    info= list(map(int, input().split()))
    u=info[0]-1
    v=info[1]-1
    l=[]
    for j in range(n):
        p=len(mat[u][j])
        for h in range(p):
            l.append(mat[u][j][h])
    l=list(set(l))
    p=len(l)
    c=0
    for j in range(p):
        co=l[j]
        lis=[False]*n
        stk=[]
        curr=u
        while(True):
            found=False
            for h in range(n):
                if(not(lis[h]) and co in mat[curr][h]):
                    stk.append(curr)
                    lis[h]=True
                    curr=h
                    found=True
                    break
            if(curr==v):
                c+=1
                break
            if(not(found)):
                if(len(stk)==0):
                    break
                curr=stk[-1]
                stk.pop(-1)
    print(c)