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. 
def arr_inp(n):
    if n == 1:
        return [int(x) for x in stdin.readline().split()]
    elif n == 2:
        return [float(x) for x in stdin.readline().split()]
    else:
        return [str(x) for x in stdin.readline().split()]


class disjointset:
    def __init__(self, n):
        self.rank, self.parent, self.n, self.nsets, self.edges = defaultdict(int), defaultdict(int), n, defaultdict(
            int), defaultdict(int)
        for i in range(1, n + 1):
            self.parent[i], self.nsets[i] = i, 1

    def find(self, x):

        if self.parent[x] == x:
            return x

        result = self.find(self.parent[x])

        self.parent[x] = result

        return result

    def union(self, x, y):
        xpar, ypar = self.find(x), self.find(y)

        # already union
        if xpar == ypar:
            self.edges[xpar] += 1
            return
        # perform union by rank
        if self.rank[xpar] < self.rank[ypar]:
            self.parent[xpar] = ypar
            self.edges[ypar] += self.edges[xpar] + 1
            self.nsets[ypar] += self.nsets[xpar]

        elif self.rank[xpar] > self.rank[ypar]:
            self.parent[ypar] = xpar
            self.edges[xpar] += self.edges[ypar] + 1
            self.nsets[xpar] += self.nsets[ypar]

        else:
            self.parent[ypar] = xpar
            self.rank[xpar] += 1
            self.edges[xpar] += self.edges[ypar] + 1
            self.nsets[xpar] += self.nsets[ypar]


from sys import stdin
from collections import defaultdict

n, m = arr_inp(1)
color = [disjointset(n) for i in range(m)]
for i in range(m):
    a, b, c = arr_inp(1)
    color[c - 1].union(a, b)

for i in range(int(input())):
    u, v = arr_inp(1)
    ans = 0
    for j in range(m):
        if color[j].find(u) == color[j].find(v):
            ans += 1
    print(ans)
