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 PHP7 (7.0.15) (judge verdict: Wrong Answer)

<?php
/*
    Problem URL : http://abc051.contest.atcoder.jp/tasks/abc051_d

    Score  :
    Result :
    Time   : ms
    Memory : KB
 */

// define('DEBUG', true);
define('DEBUG', false);

$m = array();
$ans = 0;

fscanf(STDIN, "%d %d", $N, $M);
for ($i = 0; $i < $M; $i++) {
    fscanf(STDIN, "%d %d %d", $a, $b, $c);
    $m[$a][$b] = $c;
    $m[$b][$a] = $c;
}

if (DEBUG) var_dump($m);

for ($k = 1; $k <= $M; $k++) {
    if (DEBUG) echo "\nk:{$k} ";
    for ($i = 1; $i <= $M; $i++) {
        // if ($i == $k) {
        //     if (DEBUG) echo PHP_EOL;
        //     continue;
        // }
        if (DEBUG) echo "i:{$i} ";
        for ($j = 1; $j <= $M; $j++) {
            if ($j == $i || $j == $k) {
                if (DEBUG) echo PHP_EOL;
                continue;
            }
            if (DEBUG) echo "j:{$j} ";
            if (!isset($m[$i][$j])) continue;
            if (!isset($m[$i][$k])) continue;
            if (!isset($m[$j][$k])) continue;
            // if (DEBUG) echo "root {$i} {$k} {$j}\n";
            if ($m[$i][$k] > $m[$i][$j] + $m[$j][$k]) {
                if (DEBUG) echo "X {$i} {$k} - {$i} {$j} + {$j} {$k}\n";
                $ans++;
            }
        }
    }
}

echo ($ans / 2) . PHP_EOL;