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. 
import sys

# sys.stdin = open('in')


class Dsu:
    def __init__(self):
        self.__sz = {}
        self.__parent = {}

    def parent(self, a):
        if (a not in self.__parent) or (a == self.__parent[a]):
            return a
        else:
            self.__parent[a] = self.parent(self.__parent[a])
            return self.__parent[a]

    def connect(self, a, b):
        p1 = self.parent(a)
        p2 = self.parent(b)

        if p2 != p1:
            if p1 not in self.__sz:
                self.__sz[p1] = 1
            if p2 not in self.__sz:
                self.__sz[p2] = 1

            sz1 = self.__sz[p1]
            sz2 = self.__sz[p2]

            if sz1 >= sz2:
                self.__parent[p2] = p1
                self.__sz[p1] += self.__sz[p2]
            else:
                self.__parent[p1] = p2
                self.__sz[p2] += self.__sz[p1]

    def connected(self, a, b):
        return self.parent(a) == self.parent(b)


# reading input
n, m = [int(i) for i in input().split(' ')]
g = {}
choice = {}
for j in range(m):
    a, b, c = [int(i) for i in input().split(' ')]
    if c not in g:
        g[c] = Dsu()
    g[c].connect(a, b)
    if a not in choice:
        choice[a] = []
    if b not in choice:
        choice[b] = []
    choice[a].append(c)
    choice[b].append(c)

# compute output
q = int(input())
mem = {}
for j in range(q):
    a, b = [int(i) for i in input().split(' ')]
    if a not in choice or b not in choice:
        print('0')
        continue

    if a in mem and type(mem[a]) == dict and b in mem[a]:
        s = mem[a][b]
    else:
        options = set.intersection(set(choice[a]), set(choice[b]))
        s = 0
        for c in options:
            s += int(g[c].connected(a, b))
        if a not in mem:
            mem[a] = {}
        if b not in mem:
            mem[b] = {}
        mem[a][b] = mem[b][a] = s
    print(s)