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. 
#   In this template you are not required to write code in main

import sys
inf = float("inf")

sys.setrecursionlimit(1000000)
#from cmath import sqrt
#from collections import deque, Counter, OrderedDict,defaultdict
#from heapq import nsmallest, nlargest, heapify,heappop ,heappush, heapreplace
from math import ceil,floor,log,sqrt,factorial,pow,pi,gcd
#from bisect import bisect_left,bisect_right
#import numpy as np

abc='abcdefghijklmnopqrstuvwxyz'
abd={'a': 0, 'b': 1, 'c': 2, 'd': 3, 'e': 4, 'f': 5, 'g': 6, 'h': 7, 'i': 8, 'j': 9, 'k': 10, 'l': 11, 'm': 12, 'n': 13, 'o': 14, 'p': 15, 'q': 16, 'r': 17, 's': 18, 't': 19, 'u': 20, 'v': 21, 'w': 22, 'x': 23, 'y': 24, 'z': 25}
mod,MOD=1000000007,998244353
vow=['a','e','i','o','u']
dx,dy=[-1,1,0,0],[0,0,1,-1]

class UnionFind:
    def __init__(self, n):
        self.parent = list(range(n))

    def find(self, a):  # to find the parent of a
        acopy = a
        while a != self.parent[a]:
            a = self.parent[a]
        while acopy != a:
            self.parent[acopy], acopy = a, self.parent[acopy]
        return a

    def union(self, a, b):  # to unite a and b under same object
        self.parent[self.find(b)] = self.find(a)

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

n,m=get_ints()
Arr=[0]*(105)
for i in range(105):
    Arr[i]=UnionFind(105)
for i in range(m):
    x,y,z=get_ints()
    Arr[z].union(x,y)
q=int(input())
while q>0:
    x,y=get_ints();count=0
    for i in range(105):
        if Arr[i].find(x)==Arr[i].find(y):
            count+=1
    print(count)
    q-=1