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. 
from collections import defaultdict
class DisjSet: 
    def __init__(self, n): 
        # Constructor to create and 
        # initialize sets of n items 
        self.rank = [1] * n 
        self.parent = [i for i in range(n)] 
  
  
    # Finds set of given item x 
    def find(self, x): 
          
        # Finds the representative of the set 
        # that x is an element of 
        if (self.parent[x] != x): 
              
            # if x is not the parent of itself 
            # Then x is not the representative of 
            # its set, 
            self.parent[x] = self.find(self.parent[x]) 
              
            # so we recursively call Find on its parent 
            # and move i's node directly under the 
            # representative of this set 
  
        return self.parent[x] 
  
  
    # Do union of two sets represented 
    # by x and y. 
    def Union(self, x, y): 
          
        # Find current sets of x and y 
        xset = self.find(x) 
        yset = self.find(y) 
  
        # If they are already in same set 
        if xset == yset: 
            return
  
        # Put smaller ranked item under 
        # bigger ranked item if ranks are 
        # different 
        if self.rank[xset] < self.rank[yset]: 
            self.parent[xset] = yset 
  
        elif self.rank[xset] > self.rank[yset]: 
            self.parent[yset] = xset 
  
        # If ranks are same, then move y under 
        # x (doesn't matter which one goes where) 
        # and increment rank of x's tree 
        else: 
            self.parent[yset] = xset 
            self.rank[xset] = self.rank[xset] + 1
L = []
n,m = map(int,input().split())
edges = []
for k in range(m):
  a,b,c = map(int,input().split())
  edges.append([a,b,c])
q = int(input())
queries = []
for l in range(q):
  x,y = map(int,input().split())
  queries.append([x,y])
count = [0 for i in range(q)]
for i in range(1,101):
  g = DisjSet(101)
  for [u,v,c] in edges:
    if c == i:
      g.Union(u,v)
  for z in range(q):
    if g.find(queries[z][0]) == g.find(queries[z][1]):
      count[z] += 1
for m in range(q):
    print(count[m])