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 calcs(x,y,ax,ay,bx,by) ((ax-x)*(by-y)-(bx-x)*(ay-y))/2.0 end
def inpf() a=gets.chomp.split(" ").map(&:to_f)end
def inps() a=gets.chomp.split(" ")end  
def copy(a) Marshal.load(Marshal.dump(a)) end
def kaizyo(n)(n < 2)? 1 : (2..n).inject(:*) end
def scount(a) b=na(a.max+1);a.each{|n|b[n]+=1};return b end
def na(n,d=0) Array.new(n,d)end
def na2(n,m,d=0) Array.new(n){Array.new(m,d)}end
def na3(n,m,l,d=0) Array.new(n){Array.new(m){Array.new(l,d)}}end
def bit(n) n.to_s(2).split("").map(&:to_i) end
def inp() a=gets.chomp.split(" ").map(&:to_i)end
n,m = inp
pas = na2(n+1,0)
rec = na2(n+1,n+1,999999999)
test = []
m.times do 
  a,b,c=inp
  pas[a].push([b,c])
  pas[b].push([a,c])
  test.push([a,b,c])
end
#p pas
used = na(n)
(1..n).each do |s|
  rec[s][s] = 0
  ind = s
  used = na(n+1,0)
  while(ind >= 0)  do
    max = 1e20
    nex = -1
    tab = pas[ind]
    tab.each do |a|
      hi = a[0]
      next if(used[hi] == true)
      kyori = rec[s][ind] + a[1]
      rec[s][hi] = kyori if rec[s][hi] > kyori
      if max > rec[s][hi]
        max = rec[s][hi]
        nex = hi
      end
    end
    used[ind] = true;
    ind = nex
  #rererereerer[missmissniu] += [waawawawawwa[bugbugbubg]] if(nex < 0)
  end
end
#p rec
c = 0
test.each do|a|
  aa = 0
  (1..n).each do |n|
    if((rec[a[1]][n] + a[2]) == rec[a[0]][n]) 
      aa = 1
      break 
    end
  end
  if(aa == 1)
    next
  else
    c +=1
  end
end
p c