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. 
class DisjointSet:      #only support query, NO construction supported
    def __init__(self, n):
        self.num_sets   = n         #--@union
        self.parents    = list(range(n))
        self.ranks      = [0] * n
        self.sizes      = [1] * n   #used@union: size of set@root! 0 if NOT repsentive
 
    def __str__(self):
        return '%s'%self.parents

    def __repr__(self):
        return '%s'%self.parents

    def find(self,j): #no compress
        while self.parents[j]!=j:
            j   = self.parents[j]
        return j

    def union(self,i,j):
        i   = self.find(i)
        j   = self.find(j)
        if i == j:
            return False
        rd  = self.ranks[i] - self.ranks[j]
        if   rd == 0:       # Increment repr0's rank if both nodes have same rank
            self.ranks[i] += 1
        elif rd <  0:       # Swap to ensure that repr0's rank >= repr1's rank
            i, j = j, i
        self.parents[j]  = i   #merge
        self.sizes[i]   += self.sizes[j]
        self.sizes[j]    = 0
        self.num_sets   -= 1
        return True

def main():
    n,m =  list(map(int,input().split())) #100,100
    el  = [list(map(int,input().split())) for _ in range(m)]
    q   = int(input())
    vl  = [list(map(int,input().split())) for _ in range(q)] 

    dl  = [DisjointSet(n) for _ in range(m)]
    for a,b,c in el:
        dl[c-1].union(a-1,b-1)
    [print(sum([d.find(u-1)==d.find(v-1) for d in dl])) for u,v in vl]

main()

'''
import cProfile
cProfile.run("main()")
'''
