Mr. Kitayuta has just bought an undirected graph consisting of 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 ≤ 100, 1 ≤ m ≤ 100), 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 ≤ 100), 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. 
2 1
1 2 1
1
1 2
 2 3
1 2 3
1 2 2
1 2 1
1
1 2
 2 5
1 2 1
1 2 2
1 2 3
1 2 4
1 2 5
1
1 2

1
 3
 5

#	!/usr/bin/env python3
#	coding: UTF-8
#	Modified: <19/Feb/2019 06:22:16 PM>


#	✪ H4WK3yE乡
#	Mohd. Farhan Tahir
#	Indian Institute Of Information Technology (IIIT),Gwalior

#	Question Link
#
#

# ///==========Libraries, Constants and Functions=============///


import sys

inf = float("inf")
mod = 1000000007


def get_array(): return list(map(int, sys.stdin.readline().split()))


def get_ints(): return map(int, sys.stdin.readline().split())


def input(): return sys.stdin.readline()

# ///==========MAIN=============///


N = 105
graph = [[] for _ in range(N)]
component = [0] * N
visited = [False] * N
cnt = 0
v = 0
n, m = 0, 0
ans = 0


def explore(node):
    visited[node] = True
    component[node] = cnt
    for neighbour, color in graph[node]:
        if visited[neighbour] == False:
            explore(neighbour)


def dfs(n):
    global cnt
    for i in range(1, n + 1):
        if visited[i] == False:
            cnt += 1
            explore(i)


def explore_o(node, c):
    global ans
    visited[node] = True
    for neighbour, color in graph[node]:
        if visited[neighbour] == False and color == c:
            if neighbour == v:
                ans += 1
                return
            explore_o(neighbour, c)


def dfs_o(u):
    global visited, ans
    ans = 0
    for neighbour, color in graph[u]:
        visited = [False] * N
        visited[u] = True
        if neighbour == v:
            ans += 1
            continue
        explore_o(neighbour, color)
    return ans


def main():
    global n, m
    n, m = get_ints()
    for _ in range(m):
        u, x, c = get_ints()
        graph[u].append((x, c))
        graph[x].append((u, c))
    dfs(n)
    global v
    for tc in range(int(input())):
        u, v = get_ints()
        if component[u] != component[v]:
            print(0)
            continue
        print(dfs_o(u))


if __name__ == "__main__":
    main()
