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. 
#!/usr/bin/python

from collections import deque
from collections import defaultdict as dd

def ir():
    return int(raw_input())

def ia():
    line = raw_input()
    line = line.split()
    return map(int, line)

N = 105
class CC(object):
    def __init__(self):
        self.id = [i for i in range(N)]
    def connected(self, p, q):
        return self.find(p) == self.find(q)
    def find(self, p):
        id = self.id
        while id[p]!=p:
            p = id[p]
        return p
    def union(self, p, q):
        p = self.find(p)
        q = self.find(q)
        self.id[p] = self.id[q]
    
n, m = ia()

G = dd(CC)

for i in range(m):
    a, b, c = ia()
    G[c].union(a, b)

q = ir()
for i in range(q):
    u, v = ia()
    ans = 0
    for c in G.values():
        if c.connected(u, v): ans += 1
    print ans

