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)

N, M = gets.split.map(&:to_i)
dist = Array.new(N + 1){Array.new(N + 1, Float::INFINITY)}
edges = []
direct = Array.new(N + 1){Hash.new}

M.times do |i|
    a, b, c = gets.split.map(&:to_i)
    edges.push([a, b])
    dist[a][b] = c
    dist[b][a] = c
    direct[a][b] = true
    direct[b][a] = true
end

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

(1..N).each do |i|
    (1..N).each do |j|
        (1..N).each do |k|
            d2 = dist[i][k] + dist[k][j]
            if dist[i][j] > d2
                dist[i][j] = d2
                dist[j][i] = d2
                next if i == j || j == k || k == i
                direct[i][j] = false
                direct[j][i] = false
            end
        end
    end
end

p edges.count{|arr| !direct[arr[0]][arr[1]] }