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 math,sys,bisect,heapq
from collections import defaultdict,Counter,deque
from itertools import groupby,accumulate
#sys.setrecursionlimit(200000000)
int1 = lambda x: int(x) - 1
#def input(): return sys.stdin.readline().strip()
input = iter(sys.stdin.buffer.read().decode().splitlines()).__next__
ilele = lambda: map(int,input().split())
alele = lambda: list(map(int, input().split()))
ilelec = lambda: map(int1,input().split())
alelec = lambda: list(map(int1, input().split()))
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
#MOD = 1000000000 + 7
def Y(c):  print(["NO","YES"][c])
def y(c):  print(["no","yes"][c])
def Yy(c):  print(["No","Yes"][c])

n,m = ilele()
Color = defaultdict(set)

G = defaultdict(set)    

C = set()

def addEdge(a,b):
    G[a].add(b)
    G[b].add(a)
    
def addColor(a,b,c):
    Color[(a,b)].add(c)
    Color[(b,a)].add(c)
    C.add(c)
    
for i in range(m):
    a,b,c = ilele()
    addColor(a,b,c)
    addEdge(a,b)
    
vis = [False]*(n+1)
Ans = []
    
def fun(node,dest,vis,grp):
    #print(node,vis,grp)
    if not grp:
        return
    if node == dest:
        for i in grp:
            Ans.append(i)
        return
    vis[node] = True
    for i in G.get(node,[]):
        #print(i)
        if not vis[i]:
            newvis = vis.copy()
            #newvis[c] =True
            z = grp.intersection(Color[node,i])
            #print(z)
            fun(i,dest,newvis,z)
            
#print(Color,G)
for i in range(int(input())):
    a,b = ilele()
    vis = [False]*(n+1)
    grp = C.copy()
    fun(a,b,vis,grp)
    #print(Ans)
    print(len(set(Ans)))
    Ans =[]
 

    
    
    