506_D. Mr. Kitayuta's Colorful Graph
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. 
```python

#make a table
n,m = map(int,input().split())
arr = [[] for i in range(n+1)]
for i in range(1,m+1):
    l = list(map(int,input().split()))
    arr[l[0]].append((l[1],l[2]))
    arr[l[1]].append((l[0],l[2]))


#dfs
def dfs(a,b,k):
    if a == b:
        return True
    seen.append(a)
    for i in arr[a]:
        x,c = i
        if c == k and x not in seen:
            if dfs(x,b,k):
                return True
    return False
    
query = int(input())
count = 0
score = [0]*(query)
seen = []
#get destination inputs
for i in range(query):
    a,b = map(int,input().split())
    for k in range(1,101):
        seen = []
        if dfs(a,b,k):
            count += 1
    score[i] = count
    count = 0

print("\n".join(list(map(str,score))))
```