Problem:
Score : 400 points

Problem Statement

You are given an undirected connected weighted graph with N vertices and M edges that contains neither self-loops nor double edges.

The i-th (1≤i≤M) edge connects vertex a_i and vertex b_i with a distance of c_i.

Here, a self-loop is an edge where a_i = b_i (1≤i≤M), and double edges are two edges where (a_i,b_i)=(a_j,b_j) or (a_i,b_i)=(b_j,a_j) (1≤i<j≤M).

A connected graph is a graph where there is a path between every pair of different vertices.

Find the number of the edges that are not contained in any shortest path between any pair of different vertices.

Constraints

2≤N≤100

N-1≤M≤min(N(N-1)/2,1000)

1≤a_i,b_i≤N

1≤c_i≤1000

c_i is an integer.

The given graph contains neither self-loops nor double edges.

The given graph is connected.

Input

The input is given from Standard Input in the following format:

N M
a_1 b_1 c_1
a_2 b_2 c_2
:
a_M b_M c_M

Output

Print the number of the edges in the graph that are not contained in any shortest path between any pair of different vertices.

Sample Input 1

3 3
1 2 1
1 3 1
2 3 3

Sample Output 1

1

In the given graph, the shortest paths between all pairs of different vertices are as follows:

The shortest path from vertex 1 to vertex 2 is: vertex 1 → vertex 2, with the length of 1.

The shortest path from vertex 1 to vertex 3 is: vertex 1 → vertex 3, with the length of 1.

The shortest path from vertex 2 to vertex 1 is: vertex 2 → vertex 1, with the length of 1.

The shortest path from vertex 2 to vertex 3 is: vertex 2 → vertex 1 → vertex 3, with the length of 2.

The shortest path from vertex 3 to vertex 1 is: vertex 3 → vertex 1, with the length of 1.

The shortest path from vertex 3 to vertex 2 is: vertex 3 → vertex 1 → vertex 2, with the length of 2.

Thus, the only edge that is not contained in any shortest path, is the edge of length 3 connecting vertex 2 and vertex 3, hence the output should be 1.

Sample Input 2

3 2
1 2 1
2 3 1

Sample Output 2

0

Every edge is contained in some shortest path between some pair of different vertices.

Solution in Ruby (2.3.3) (judge verdict: Wrong Answer)

n,m=gets.split

n=n.to_i
m=m.to_i


array=[]
m.times{
    
    array<<gets.split.map(&:to_i)}
    
array2=Array.new(n).map{Array.new(n,10**10)}
i=0
n.times{
    array2[i][i]=0
    i+=1
    }
    
j=0
m.times{
  array2[array[j][0]-1][array[j][1]-1]=array[j][2]
  array2[array[j][1]-1][array[j][0]-1]=array[j][2]
  
    j+=1
    
}
 
n/2.times{
    j=array2.clone
    
 s=0
n.times{
    t=0
    n.times{
       u=0 
        n.times{
            
            array2[s][t]=[array2[s][u]+array2[u][t],array2[s][t]].min
            
        u+=1    
        }
        t+=1
        }
    
   s+=1 
}



}


answer=0
kk=0
m.times{
    if array2[array[kk][0]-1][array[kk][1]-1]!=array[kk][2]
        answer+=1
    end
    
    kk+=1
    
}



puts answer