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)

#include<iostream>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<vector>
#include<deque>
using namespace std;
#define rep(i,s,n) for(int i = s;i<n;i++)
#define repe(i,s,n) for(int i = s;i<=n;i++)
#define pb push_back
#define fi first
#define se second
typedef long long ll;
typedef pair<int,int>pint;
typedef vector<int>vint;
typedef vector<pint>vpint;
static const ll maxLL = (ll)1 << 62;
const int mod=1000000007;
int dy[]={-1,0,1,0};
int dx[]={0,1,0,-1};
int dyy[]={0,-1,0,1};
int dxx[]={-1,0,1,0};

//#define int ll

int n,m;
int a[1001],b[1001],c[1001];
int d[1001][1001];
int dd[1001][1001];
int cnt;

void warshall_floyd(int n){
  rep(i,0,n)rep(j,0,n)rep(k,0,n){
    if(d[j][k]>d[j][i]+d[i][k] )d[j][k]=d[j][i]+d[i][k];
  }
}

signed main(){
    cin>>n>>m;
    rep(i,0,1000)rep(j,0,1000){
        if(i!=j)d[i][j]=1<<28;
        else d[i][j]=0;
    }
    rep(i,0,m){
        cin>>a[i]>>b[i]>>c[i];
        a[i]--,b[i]--;
        d[a[i]][b[i]]=c[i];
        d[b[i]][a[i]]=c[i];
        dd[a[i]][b[i]]=dd[b[i]][a[i]]=c[i];
    }
    warshall_floyd(n);
    
    cnt=0;
    rep(i,0,n){
        rep(j,i+1,n){
            if(dd[i][j]>d[i][j])cnt++;
        }
    }
    
    cout<<cnt<<endl;
    
    return 0;
}