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: Accepted)

N,M = gets.split.map &:to_i
INF = 10**15
ds = (0...N).map{ Array.new(N, INF) }
es = (0...N).map{ (0...N).map{ [] } }
M.times {
  a,b,c = gets.split.map &:to_i
  a -= 1; b -= 1
  ds[a][b] = ds[b][a] = c
  es[a][b] = [b]; es[b][a] = [a]
}

(0...N).each{ |k|
  dsk = ds[k]
  esk = es[k]
  (0...N).each{ |i|
    next if i == k
    dsi = ds[i]
    esi = es[i]
    (0...N).each{ |j|
      next if i == j
      next if k == j
      v = dsi[k] + dsk[j]
      if v < dsi[j]
        dsi[j] = v
        esi[j] = esi[k] + esk[j]
      end
    }
  }
}

used = {}
(0...N).each{ |i|
  dsi = ds[i]
  esi = es[i]
  (0...N).each{ |j|
    next if dsi[j] == INF
    arr = [i] + esi[j]
    (0...arr.size-1).each{ |a|
      used[arr[a]*N+arr[a+1]] = 1
    }
  }
}
p M - used.size/2