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: Wrong Answer)

#include"bits/stdc++.h"
#define int long long
#define _overload3(_1,_2,_3,name,...) name
#define _rep(i,n) repi(i,0,n)
#define repi(i,a,b) for(int i=(a);i<(b);++i)
#define rep(...) _overload3(__VA_ARGS__,repi,_rep,)(__VA_ARGS__)
#define itr(i,x) for(auto i=(x).begin();i!=(x).end();++i)
#define All(x) (x).begin(),(x).end()
#define gridif(a,x,b) if((a)<=(x)&&(x)<(b))

using namespace std;
typedef pair<int,int> P;
const int mod=1000000007;

int digitsum(int n,int b){
	if(b<2) return -1;
	if(n<b) return n;
	return digitsum(n/b,b)+n%b;
}
int mpow(int a,int x);
int m_inv(int n);
vector<int> split(int n,int a);
string xal_number(int n,int x);
int gcd(int x,int y){return y?gcd(y,x%y):x;}
int lcm(int x,int y){return x*y/gcd(x,y);}
class Factorial{
private:
	vector<int> fac;
public:
	Factorial(int N){
		fac.push_back(1);
		rep(i,N) fac.push_back(fac[i]*(i+1)%mod);
	}
	int fact(int a){return fac[a];}
	int ifac(int a){return m_inv(fac[a]);}
	int cmb(int n,int r);
};
struct UnionFind {
    vector<int> par; // 親ノード
    UnionFind(int n = 1) { init(n);}
    void init(int n = 1) {
        par.resize(n);
        rep(i,n) par[i] = -1;
    }
    int root(int x) {
        if (par[x] <0) return x;
        else return par[x] = root(par[x]);      
    }
    int size(int x){ return -par[root(x)];}
    bool issame(int x, int y) {return root(x)==root(y);}
    bool connect(int x, int y);
};

signed main(){
	int N,M;
	cin>>N>>M;
	int INF=1<<29;
	vector<int> a(M),b(M);
	vector<vector<int>> dist(N,vector<int>(N,INF));
	rep(i,M){
		int c;
		cin>>a[i]>>b[i]>>c;
		a[i]--;b[i]--;
		dist[a[i]][b[i]]=c;
		dist[b[i]][a[i]]=c;
	}
	rep(k,N) rep(i,N) rep(j,N){
		if(dist[i][j]>dist[i][k]+dist[k][j]) dist[i][j]=INF;
	}
	int ans=0;
	rep(i,M) if(dist[a[i]][b[i]]==INF) ans++;
	cout<<ans<<endl;
}