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)

#ifdef LOCAL
    #define _GLIBCXX_DEBUG
    #define __clock__
#else
    #pragma GCC optimize("Ofast")
#endif
#include<bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
using VI = vector<ll>;
using VV = vector<VI>;
using VS = vector<string>;
using PII = pair<ll, ll>;

// tourist set
template <typename A, typename B>
string to_string(pair<A, B> p);

template <typename A, typename B, typename C>
string to_string(tuple<A, B, C> p);

template <typename A, typename B, typename C, typename D>
string to_string(tuple<A, B, C, D> p);

string to_string(const string& s) {
  return '"' + s + '"';
}

string to_string(const char* s) {
  return to_string((string) s);
}

string to_string(bool b) {
  return (b ? "true" : "false");
}

string to_string(vector<bool> v) {
  bool first = true;
  string res = "{";
  for (int i = 0; i < static_cast<int>(v.size()); i++) {
    if (!first) {
      res += ", ";
    }
    first = false;
    res += to_string(v[i]);
  }
  res += "}";
  return res;
}

template <size_t N>
string to_string(bitset<N> v) {
  string res = "";
  for (size_t i = 0; i < N; i++) {
    res += static_cast<char>('0' + v[i]);
  }
  return res;
}

template <typename A>
string to_string(A v) {
  bool first = true;
  string res = "{";
  for (const auto &x : v) {
    if (!first) {
      res += ", ";
    }
    first = false;
    res += to_string(x);
  }
  res += "}";
  return res;
}

template <typename A, typename B>
string to_string(pair<A, B> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}

template <typename A, typename B, typename C>
string to_string(tuple<A, B, C> p) {
  return "(" + to_string(get<0>(p)) + ", " + to_string(get<1>(p)) + ", " + to_string(get<2>(p)) + ")";
}

template <typename A, typename B, typename C, typename D>
string to_string(tuple<A, B, C, D> p) {
  return "(" + to_string(get<0>(p)) + ", " + to_string(get<1>(p)) + ", " + to_string(get<2>(p)) + ", " + to_string(get<3>(p)) + ")";
}

void debug_out() { cerr << '\n'; }

template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << to_string(H);
  debug_out(T...);
}

#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 42
#endif
// tourist set end

template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }

#define FOR(i,a,b) for(ll i=(a);i<(b);++i)
#define rep(i,b) FOR(i, 0, b)
#define ALL(v) (v).begin(), (v).end()
#define p(s) cout<<(s)<<'\n'
#define p2(s, t) cout << (s) << " " << (t) << '\n'
#define br() p("")
#define pn(s) cout << (#s) << " " << (s) << '\n'
#define SZ(x) ((int)(x).size())
#define SORT(A) sort(ALL(A))
#define RSORT(A) sort(ALL(A), greater<ll>())
#define MP make_pair
#define p_yes() p("Yes")
#define p_no() p("No")

ll SUM(VI& V){
  return accumulate(ALL(V), 0LL);
}

ll MIN(VI& V){return *min_element(ALL(V));}
ll MAX(VI& V){return *max_element(ALL(V));}

void print_vector(VI& V){
  ll n = V.size();
  rep(i, n){
    if(i) cout << ' ';
    cout << V[i];
  }
  cout << endl;
}

ll gcd(ll a,ll b){
    if(b == 0) return a;
    return gcd(b,a%b);
}

ll lcm(ll a,ll b){
    ll g = gcd(a,b);
    return a / g * b;
}

// long double
using ld = long double;
#define EPS (1e-14)
#define equals(a,b) (fabs((a)-(b)) < EPS)

void no(){p_no(); exit(0);}
void yes(){p_yes(); exit(0);}

const ll mod = 1e9 + 7;
const ll inf = 1e18;
const double PI = acos(-1);

struct WarshallFloyd{
  VV d;
  ll N;
  WarshallFloyd(ll n){
    N = n;
    d.resize(N, VI(N, inf));
    rep(i, N) d[i][i] = 0;
  }
  // 双方向
  void register_edge2(ll a, ll b, ll c){
    d[a][b] = c;
    d[b][a] = c;
  }
  void calc(){
    rep(i, N){ // 経由点
      rep(j, N){ // 始点
         rep(k, N){ // 終点
          d[j][k] = min(d[j][k], d[j][i] + d[i][k]);
        }
      }
    }
  }
  ll distance(ll a, ll b){
    return d[a][b];
  }
};

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

    // input
    ll N,M;
    cin>>N>>M;

    VI A(M);
    VI B(M);
    VI C(M);
    rep(i,M){
      cin>>A[i]>>B[i]>>C[i];
      A[i]--;
      B[i]--;
    }

    WarshallFloyd wa(N);

    rep(i,M){
      wa.register_edge2(A[i],B[i],C[i]);
    }
    wa.calc();

    ll cnt=0;
    rep(i,M){
      ll a = A[i];
      ll b = B[i];
      if(wa.distance(a,b)<C[i])cnt++;
    }
    p(cnt);


    return 0;
}