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 C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Linq;
using System.Collections.Generic;
class P
{
    static void Main()
    {
        int[] nm = Console.ReadLine().Split().Select(int.Parse).ToArray();
        int[][] path = Enumerable.Repeat(0, nm[1]).Select(_ => Console.ReadLine().Split().Select(int.Parse).ToArray()).ToArray();

        int[][] map = Enumerable.Repeat(0, nm[0]).Select(_ => Enumerable.Repeat(int.MaxValue / 2, nm[0]).ToArray()).ToArray();
        for (int i = 0; i < path.Length; i++)
        {
            map[path[i][0] - 1][path[i][1] - 1] = path[i][2];
            map[path[i][1] - 1][path[i][0] - 1] = path[i][2];
        }

        bool flag = true;
        while (flag)
        {
            flag = false;
            for (int i = 0; i < nm[0]; i++)
            {
                for (int j = 0; j < nm[0]; j++)
                {
                    if (i != j)
                    {
                        for (int k = 0; k < nm[0]; k++)
                        {
                            if (k != i && k != j && map[i][j] > map[i][k] + map[k][j])
                            {
                                flag = true;
                                map[i][j] = map[i][k] + map[k][j];
                                map[j][i] = map[i][k] + map[k][j];
                            }
                        }
                    }
                }
            }
        }

        int res = 0;
        for (int i = 0; i < path.Length; i++)
        {
            if (map[path[i][0] - 1][path[i][1] - 1] < path[i][2])
            {
                res++;
            }
        }
        Console.WriteLine(res);
    }
}