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)

def saiki(x,md,d,fx,fd)
  fd.push(x)
  ans = 0
    $keiro[x].each do |xx|
      next if (fd.include?(xx[0]))
      return ans + 0 if(d > md)
      if(xx[0] == fx)
        if(d + xx[1] < md)
          return ans + 1
        else
          return ans
        end
      else
      fd = Marshal.load(Marshal.dump(fd))
      ans += saiki(xx[0],md,d+xx[1],fx,fd)
      end
    end
  return ans
end
ans = 0
$h = {}
n,k = gets.chomp.split(" ").map(&:to_i)
a = []
$keiro = Array.new(n+1){Array.new(0)}
k.times do
  b = (gets.chomp.split(" ").map(&:to_i))
  $keiro[b[0]].push([b[1],b[2]])
  $keiro[b[1]].push([b[0],b[2]])
  a.push(b)
end
a.each do |aa|
 ($keiro[aa[0]].sort{|a,b|a[1] <=> b[1]})
 next if($keiro[aa[0]][0][1] == aa[1])
 if(saiki(aa[0],aa[2],0,aa[1],[]) > 0)
 ans += 1
 end
end
puts ans