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)
a = []
b = []
c = []
for i in 0..M-1 do 
    a[i],b[i],c[i] = gets.split.map(&:to_i)
end
dist = Array.new(N).map{Array.new(N)}
#distの初期化
for i in 0..N-1 do
    for j in 0..N-1 do
        if i == j then
            dist[i][j] = 0
        else
            dist[i][j] = 10000
        end
    end
end

for i in 0..M-1 do
    dist[a[i]-1][b[i]-1] = c[i]
    dist[b[i]-1][a[i]-1] = c[i]
end
#ワーシャルフロイド法
for k in 0..N-1 do
    for i in 0..N-1 do
        for j in 0..N-1 do
            dist[i][j] = [dist[i][j],dist[i][k] + dist[k][j]].min
        end
    end
end
#使用されていない辺の計算
sum = 0
for i in 0..M-1 do
    if dist[a[i]-1][b[i]-1] < c[i] then
        sum = sum + 1
    end
end
puts sum