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 ufds:
    def __init__(self):
        self.rank = dict()
        self.value = dict()

    def union(self,a,b):
        x,y = self.findSet(a),self.findSet(b)
        if self.rank[x] > self.rank[y]:
            self.value[y] = x
        else:
            self.value[x] = y
        if self.rank[x] == self.rank[y]:
            self.rank[y] += 1
            
    def findSet(self,x):#Path Compression
        if x != self.value[x]:
            self.value[x] = self.findSet(self.value[x])
        return self.value[x]
            
    def sameSet(self,a,b):
        return self.findSet(a) == self.findSet(b)
    
    def createKey(self,x):
        self.rank.update({x:0})
        self.value.update({x:x})
    
a = list(map(int,input().split()))
size = a[0]
graph = dict()
for i in range(a[1]):
    b = list(map(int,input().split()))
    color = b[2]
    if color not in graph.keys():
        graph.update({color:ufds()})
        for i in range(size):
            graph[color].createKey(i+1)
    graph[color].union(b[0],b[1])
for i in range(int(input())):
    b = list(map(int,input().split()))
    counter = 0
    for j in graph.keys():
        if graph[j].sameSet(b[0],b[1]):
            counter+=1
    print(counter)