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 PHP (7.4.4) (judge verdict: Accepted)

<?php
  list($n, $m) = explode(" ", trim(fgets(STDIN)));

  $tree = [];
  $tree2 = [];
  for($i=1; $i<=$n; $i++){
    $tree2[$i] = [];
    for($j=1; $j<=$n; $j++){
      if($i == $j){
        $tree2[$i][$j] = 0;
      }else{
        $tree2[$i][$j] = PHP_INT_MAX;
      }
    }
  }
  for($i=0; $i<$m; $i++){
    list($a, $b, $c) = explode(" ", trim(fgets(STDIN)));
    if(!isset($tree[$a])) $tree[$a] = [];
    $tree[$a][$b] = $c;
    $tree2[$a][$b] = $c;
    $tree2[$b][$a] = $c;
  }

  for($k=1; $k<=$n; $k++){
    for($i=1; $i<=$n; $i++){
      for($j=1; $j<=$n; $j++){
        $tree2[$i][$j] = min($tree2[$i][$j], $tree2[$i][$k]+$tree2[$k][$j]);
      }
    }
  }

  $ans = 0;

  foreach($tree as $i => $t){
    foreach($t as $j => $v){
      if($v != $tree2[$i][$j]){
//        echo $i." ".$j." ".$v." ".$tree2[$i][$j].PHP_EOL;
        $ans++;
      }
    }
  }

  echo $ans;