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: Runtime Error)

//#define _GLIBCXX_DEBUG
#include<bits/stdc++.h>
#define PI 3.14159265359
using namespace std;
#define rep(i, n) for (ll i = 0; i < (ll)(n); i++)
const long long INF= 1e+18+1;
typedef long long ll;
typedef vector<ll> vl;
typedef vector<vector<ll> >vvl;
typedef pair<ll,ll> P;
typedef tuple<ll,ll,ll> T;
const ll MOD=1000000007LL;
string abc="abcdefghijklmnopqrstuvwxyz";
string ABC="ABCDEFGHIJKLMNOPQRSTUVWXYZ";
struct Edge {
    long long to;
    long long cost;
};
using Graph = vector<vector<Edge>>;
void dijkstra(const Graph &G,ll s, vector<ll> &dis, vector<ll> &prev) {
    ll N = G.size();
    dis.resize(N, INF);
    prev.resize(N, -1); // 初期化
    priority_queue<P, vector<P>, greater<P>> pq; 
    dis[s] = 0;
    pq.emplace(dis[s], s);
    while (!pq.empty()) {
        P p = pq.top();
        pq.pop();
        ll v = p.second;
        if (dis[v] < p.first) {
            continue;
        }
        for (auto &e : G[v]) {
            if (dis[e.to] > dis[v] + e.cost) {
                dis[e.to] = dis[v] + e.cost;
                prev[e.to] = v; // 頂点 v を通って e.to にたどり着いた
                pq.emplace(dis[e.to], e.to);
            }
        }
    }
}
int main(){
  ll n,m;cin>>n>>m;
  vector<vector<bool> >data(n,vector<bool>(n,false));
  vector<vector<Edge> > G;
  rep(i,m){
    ll a,b,c;cin>>a>>b>>c;
    a--;b--;
    data[a][b]=true;
    data[b][a]=true;
    Edge e;
    e.to=b;e.cost=c;
    G[a].push_back(e);
  }
  rep(i,n){
    vl prev;
    vl dis;
    dijkstra(G,i,dis,prev);
    rep(j,prev.size()-1){
      data[prev[j]][prev[j+1]]=false;
      data[prev[j+1]][prev[j]]=false;
    }
  }
  ll count=0;
  rep(i,n){
    rep(j,n){
      if(data[i][j])count++;
    }
  }
  cout<<count/2<<endl;
}