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
from sys import stdin, stdout

par = defaultdict(list)

def build(n):
    global par, size
    for c in range(101):
        par[c] = []
        for i in range(n):
            par[c].append(i)
    return par

def find(c, i):
    global par
    if par[c][i] == i:
        return i
    par[c][i] = find(c, par[c][i])
    return par[c][i]

def union(a, b, c):
    global par, scc, size
    p = find(c, a)
    q = find(c, b)
    if p == q:
        return
    par[c][q] = par[c][p]

def main():
    (n, m) = map(int, stdin.readline().strip().split(' '))
    par = build(n)
    max_c = 0
    for i in range(m):
        (a, b, c) = map(lambda i: int(i) - 1, stdin.readline().strip().split(' '))
        union(a, b, c)
        max_c = max(max_c, c)
    q = int(stdin.readline().strip())
    for i in range(q):
        (a, b) = map(lambda i: int(i) - 1, stdin.readline().strip().split(' '))
        count = 0
        for c in range(max_c + 1):
            p = find(c, a)
            q = find(c, b)
            if p == q:
                count += 1
        stdout.write('{}\n'.format(count))
main()
