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.strip.split(' ').map(&:to_i)

dist = Array.new(n+1) { [10**12]*(n+1) }

(1..n).each do |i|
  dist[i][i]=0
end

abc = readlines.map {|line| line.strip.split.map(&:to_i) }

abc.each do |a, b, c|
  dist[a][b] = c
  dist[b][a] = c
end

# ここまで隣接行列の初期化処理
# dist[a][b]には頂点a,b間の辺のコストを入れておく
# dist[a][a]には0、辺が存在しない場合には 10**12 (INFINITY) を入れておく

# warshall floyd
(1..n).each do |via|    # 経由する頂点
  (1..n).each do |from| # 始点
    (1..n).each do |to| # 終点
      dist[to][from] = dist[from][to] = [dist[from][to], dist[from][via] + dist[via][to]].min
    end
  end
end
 
result = 0
abc.each do |a, b, c|
  # 2つの頂点を結ぶ辺が最短でなければいずれの最短経路からも利用されることはない
  result += 1 if dist[a][b] < c
end
 
puts result