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>
//#pragma comment(linker, "/STACK:102400000,102400000")
#include <stdio.h>
#include <iostream>
#include <iomanip>
#include <fstream>
#include <string.h>
#include <stdlib.h>
#include <time.h>
#include <limits.h>
#include <math.h>
#include <algorithm>
#include <deque>
#include <queue>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <utility>
#include <sstream>
#include <complex>
#include <string>
#include <vector>
#include <bitset>
#include <functional>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef double db;
typedef long double ldb;
#define rep(i,a,b) for(i=a;i<b;i++)
#define repu(i,a,b) for(i=a;i<=b;i++)
#define repd(i,a,b) for(i=a;i>=b;i--)
#define repc(i,a,b,c) for(i=a;i<b;i+=c)
#define repcu(i,a,b,c) for(i=a;i<=b;i+=c)
#define repcd(i,a,b,c) for(i=a;i>=b;i-=c)
#define FIN freopen("input.in","r",stdin)
#define FOUT freopen("output.out","w",stdout)
#define CC(a) memset(a,0,sizeof(a))
#define MST(a,b) memset(a,b,sizeof(a))
#define PII pair<int,int>
#define PLL pair<long long,long long>
#define VI vector<int>
#define VL vector<long long>
#define VII vector<PII>
#define VLL vector<PLL>
#define all(x) x.begin(),x.end()
#define mp make_pair
#define pb push_back
#define pf push_front
 
template<typename T>inline void gn(T &x){
	x=0;T f=1;char ch=getchar();
	while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
	while(isdigit(ch)){x=x*10+ch-'0';ch=getchar();}
	x*=f;
}
template<typename T>inline void gt(T x){
	if(!x)return;
	gt(x/10);putchar(x%10+'0');
}
 
const LL MAXN=1e2+5;
LL t,n,m,q,ans;
LL MOD=1e9+7,inf=0x1f1f1f1f;
db eps=1e-5;

LL f[MAXN][MAXN],a[MAXN],b[MAXN],c[MAXN];

int main()
{
//	ios_base::sync_with_stdio(false);cin.tie(0);
	LL i,j,k,l,ca=0,u,v,w;
	cin>>n>>m;
	repu(i,1,n)repu(j,1,n)f[i][j]=(i==j?0:inf);
	repu(i,1,m){
		cin>>a[i]>>b[i]>>c[i];
		f[a[i]][b[i]]=min(f[a[i]][b[i]],c[i]);
		f[b[i]][a[i]]=min(f[b[i]][a[i]],c[i]);
	}
	repu(i,1,n) repu(j,1,n) repu(k,1,n)
		f[i][j]=min(f[i][j],f[i][k]+f[k][j]);
	
	repu(j,1,m){
		k=0;
		repu(i,1,n){
			if(f[i][b[j]]<f[i][a[j]]+c[j])
				k++;
		}
		ans+=k==n;
	}
	
	cout<<ans<<endl;
	return 0;
}