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<bits/stdc++.h>
using namespace std;
#define ALL(obj) (obj).begin(), (obj).end()
#define REP(i, n) for(int i = 0;i < (n);++i)

#define endl '\n'
#define pb emplace_back
#define mkp make_pair
#define p_q priority_queue
#define INF 1000000000
#define PI 3.1415926
#define coutdb cout << fixed << setprecision(10)

const int dx[]={1,0,-1,0}, dy[]={0,-1,0,1};
const int dx_8[]={1,1,0,-1,-1,-1,0,1}, dy_8[]={0,-1,-1,-1,0,1,1,1};
const long long MOD = 1e9+7;

typedef long long ll;
// #define int long long

template <typename T> inline T GCD(T a,T b){T c;while(b!=0){c=a%b;a=b;b=c;}return a;}
template <typename T> inline T LCM(T a,T b){T c=GCD(a,b);a/=c;return a*b;}
template <typename T> inline T nCr(T a,T b){T i,r=1;for(i=1;i<=b;i++){r*=(a+1-i);r/=i;}return r;}
template <typename T> inline T nHr(T a,T b){return nCr(a+b-1,b);}

bool need[105][105];

signed main(void){
  cin.tie(0);
  ios::sync_with_stdio(false);

  int N, M, a[1000], b[1000], c[1000]; cin >> N >> M;
  int d[105][105];
  REP(i, 105) REP(j, 105) {
    if(i == j) d[i][j] = 0;
    else d[i][j] = INF;
  }
  REP(i, M){
    cin >> a[i] >> b[i] >> c[i];
    a[i]--; b[i]--;
    d[a[i]][b[i]] = d[b[i]][a[i]] = c[i];
  }
  REP(k, N) REP(i, N) REP(j, N) d[i][j] = min(d[i][j], d[i][k]+d[k][j]);

  int ans = M;
  REP(i, M){
    bool shortest = 0;
    REP(s, N) REP(t, N) if(d[s][a[i]]+c[i]+d[b[i]][t] == d[s][t] or d[s][b[i]]+c[i]+d[a[i]][t] == d[s][t]) shortest = 1;
    if(shortest) ans--;
  }
  cout << ans << endl;

  return 0;
}