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)

class Array
  def warshall_floyd
    v = size

    used = Array.new(v) { Array.new(v, false) }
    dist = map(&:dup)

    v.times do |k|
      v.times do |i|
        v.times do |j|
          cost = dist[i][k] + dist[k][j]

          if dist[i][j] > cost
            used[i][k] = used[k][i] = true
            used[j][k] = used[k][j] = true
            used[i][j] = used[j][i] = false
            dist[i][j] = cost
          end
        end
      end
    end

    used
  end
end

N, M = gets.split.map(&:to_i)
graph = Array.new(N) { Array.new(N, Float::INFINITY) }

N.times do |i|
  graph[i][i] = 0
end

edges = M.times.map { gets.split.map(&:to_i) }

edges.each do |a, b, c|
  graph[a - 1][b - 1] = c
  graph[b - 1][a - 1] = c
end

used = graph.warshall_floyd

ans = 0

puts edges.count { |a, b, _| !used[a - 1][b - 1] }