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 (5.6.30) (judge verdict: Accepted)

<?php
fscanf(STDIN, "%d %d",$N,$M);
for ($i=1;$i<=$M;$i++){
	fscanf(STDIN, "%d %d %d",$a[$i],$b[$i],$c[$i]);
}

//頂点間の距離を無限で初期化
for ($i=1;$i<=$N;$i++){
	for ($j=1;$j<=$N;$j++){
		if ($i==$j){
			$dist[$i][$j] = 0;
		}else {
			$dist[$i][$j] = INF;
		}
	}
}

//既知の経路
for ($i=1;$i<=$M;$i++){
	$dist[$a[$i]][$b[$i]] = $c[$i];
    $dist[$b[$i]][$a[$i]] = $c[$i];
}

//最短経路判定
for ($k=1;$k<=$N;$k++){
	for ($i=1;$i<=$N;$i++){
		for ($j=1;$j<=$N;$j++){
			//$iから$jに向かう場合$kを経由した方が早いかの判定
			if ($dist[$i][$k]+$dist[$k][$j]<$dist[$i][$j]){
				$dist[$i][$j] = $dist[$i][$k]+$dist[$k][$j];
                $dist[$j][$i] = $dist[$i][$k]+$dist[$k][$j];
			}
		}
	}
}

//最短経路に含まれない道をカウント
$ans = 0;
for ($i=1;$i<=$M;$i++){
	if ($dist[$a[$i]][$b[$i]] < $c[$i]){
		$ans++;
	}
}
echo $ans;
//print_r($dist);