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++14 (GCC 5.4.1) (judge verdict: Accepted)

#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <cmath>
#include <algorithm>
#include <vector>
#include <numeric>
#include <queue>
#include <map> 
#include <set>
#include <string>
#include <functional>
#include <list>
#include <random>
#include <time.h>
#define int long long
#define oku7 1000000007
#define MAXN (int)1e+5 * 2+1
#define LL_MAX 9223372036854775807	//ない環境用
using namespace std;
std::mt19937 mt((int)time(0));

int dx[4] = { 0, 1, 0,-1 }; // x軸方向への変位
int dy[4] = { 1, 0,-1, 0 }; // y軸方向への変位



int G[101][101];
int cou[101][101];



signed main() {
	int N, M;
	cin >> N >> M;
	for (int i = 0; i < 101; i++) {
		for (int j = 0; j < 101; j++) {
			if (i == j) G[i][j] = 0;
			else G[i][j] = LL_MAX/4;
		}
	}

	for (int i = 0; i < M; i++) {
		int a, b, c;
		cin >> a >> b >> c;
		G[a][b] = c;
		G[b][a] = c;
		cou[a][b] = 1;
		cou[b][a] = 1;
	}

	//ワーシャルフロイドをする、更新があって、それがLL_MAX/2でない場合はそれが使わない辺
	int ans = 0;
	for (int k = 1; k <= N; k++) {
		for (int i = 1; i <= N; i++) {
			for (int j = 1; j <= N; j++) {
				if (G[i][j] > G[i][k] + G[k][j]) {
					if (cou[i][j] == 1) {
						ans++;
						cou[i][j]--;
					}
					G[i][j] = G[i][k] + G[k][j];

				}
			}
		}
	}
	
	cout << ans/2 << "\n";

	return 0;




}