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. 
l=map(int,raw_input().split())
dict={}
count=[0]


for i in range(l[1]):
    edge=map(int,raw_input().split())
    if edge[2] in dict:
        dict[edge[2]].append([edge[0],edge[1]])
    else:
        dict[edge[2]]=[[edge[0],edge[1]]]
z=input()

for k in dict.keys():
    adj_list=[]
    for i in range(l[0]+1):
        adj_list.append([])
    for v in dict[k]:
        adj_list[v[0]].append(v[1])
        adj_list[v[1]].append(v[0])
    dict[k]=adj_list


def search_DFS(G,s,d):
    if(colour[s]==0):
        colour[s]=1
    for i in G[s]:
        if(colour[i]==0):
            colour[i]=1
            if(i==d):
                count[0]+=1
                return None
            search_DFS(G,i,d)


for j in range(z):
    go=map(int,raw_input().split())
    for k in dict.keys():
        colour=[]
        for i in range(l[0]+1):
            colour.append(0)
        search_DFS(dict[k],go[0],go[1])
    print count[0]
    count[0]=0
