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 Solution:
    def __init__(self, n, m, edges):
        self.n = n
        self.m = m
        self.edges = [[[] for _ in range(m+1)] for _ in range(n+1)]
        for _from, _to, _color in edges:
            self.edges[_from][_color].append(_to)
            self.edges[_to][_color].append(_from)
        self.id = 0

    def initialize(self):
        self.discovered = [[0]*(self.n+1) for i in range(self.m+1)]

    def bfs(self, x0, color):
        self.id += 1
        stack = [x0]
        while stack:
            n = stack.pop()
            if not self.discovered[color][n]:
                self.discovered[color][n] = self.id
                for neigh in self.edges[n][color]:
                    stack.append(neigh)

def solve(graph, colors, u, v):
    counter = 0
    for color in colors:
        if not graph.discovered[color][u] and not graph.discovered[color][v]:
            graph.bfs(u, color)
        if graph.discovered[color][u] == graph.discovered[color][v]:
            counter += 1
    return counter

def main():
    n, m = map(int, input().split())
    edges = []
    colors = set()
    for _ in range(m):
        a, b, c = map(int, input().split())
        colors.add(c)
        edges.append((a, b, c))
    graph = Solution(n, m, edges)
    graph.initialize()
    q = int(input())
    for _ in range(q):
        u, v = map(int, input().split())
        print(solve(graph, colors, u, v))

if __name__ == '__main__':
    main()
        

    
        
