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

import sys, os.path
from collections import*
from copy import*
import math
mod=10**9+7
if(os.path.exists('input.txt')):
    sys.stdin = open("input.txt","r")
    sys.stdout = open("output.txt","w")   

def dfs(x,y,z):
    vis[x]=1
    if(x==y):
        return
    for v in adj[x]:
        if(vis[v[0]]):
            continue
        if(v[1]==z):
            dfs(v[0],y,z)

adj=defaultdict(list)
n,m=map(int,input().split())
s=set()
for i in range(m):
    a,b,c=map(int,input().split())
    adj[a].append((b,c))
    adj[b].append((a,c))
    s.add(c)
for i in adj:
    print(i,adj[i])
print(s)
q=int(input())
for i in range(q):
    x,y=map(int,input().split())
    count=0
    for j in s:
        vis=[0 for i in range(n+1)]
        dfs(x,y,j)
        if(vis[y]):
            count+=1
    print(count)