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
#define _USE_MATH_DEFINES
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <string>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <functional>
#include <utility>
#include <tuple>
#include <cctype>
using namespace std;
#define INF 0x3f3f3f3f
//#define INF 1100000000000000000LL
#define MOD 1000000007
#define mp make_pair
#define mt make_tuple
#define pb push_back
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pint;
typedef tuple<int,int,int> tint;
typedef vector<int> vint;
typedef vector<ll> vll;
int dx[8]={0,0,-1,1,1,1,-1,-1};
int dy[8]={-1,1,0,0,1,-1,1,-1};
const int SIZE=200;
//ここまでテンプレ
int main() {
	//頂点数、辺数を受け取る
	int N,M;
	cin>>N>>M;
	vector<tint> E;
	//最短距離行列
	int d[SIZE][SIZE]={};
	//行列の初期化
	for(int i=0;i<N;i++){
		for(int j=0;j<N;j++){
			if(i==j)
				d[i][j]=0;
			else
				d[i][j]=INF;
		}
	}
	//辺の情報を受け取る
	//ノード番号は0-indexedに直す
	for(int i=0;i<M;i++){
		int a,b,c;
		cin>>a>>b>>c;
		a--;b--;
		d[a][b]=c;
		d[b][a]=c;
		E.pb(mt(a,b,c));
	}
	//ワーシャルフロイド
	for(int k=0;k<N;k++){
		for(int i=0;i<N;i++){
			for(int j=0;j<N;j++){
				d[i][j]=min(d[i][j],d[i][k]+d[k][j]);
			}
		}
	}
	int use=0;
	for(int i=0;i<M;i++){
		int a,b,c;
		tie(a,b,c)=E[i];
		bool flag=0;
		for(int j=0;j<N;j++){
			if(d[j][a]+c==d[j][b]){
				flag=1;
				break;
			}
		}
		if(flag)
			use++;
	}
	cout<<M-use<<endl;
    return 0;
}