Problem:
Score : 300 points

Problem Statement

Takahashi has N balls. Initially, an integer A_i is written on the i-th ball.

He would like to rewrite the integer on some balls so that there are at most K different integers written on the N balls.

Find the minimum number of balls that Takahashi needs to rewrite the integers on them.

Constraints

1 \leq K \leq N \leq 200000

1 \leq A_i \leq N

All input values are integers.

Input

Input is given from Standard Input in the following format:

N K
A_1 A_2 ... A_N

Output

Print the minimum number of balls that Takahashi needs to rewrite the integers on them.

Sample Input 1

5 2
1 1 2 2 5

Sample Output 1

1

For example, if we rewrite the integer on the fifth ball to 2, there are two different integers written on the balls: 1 and 2.
On the other hand, it is not possible to rewrite the integers on zero balls so that there are at most two different integers written on the balls, so we should print 1.

Sample Input 2

4 4
1 1 2 2

Sample Output 2

0

Already in the beginning, there are two different integers written on the balls, so we do not need to rewrite anything.

Sample Input 3

10 3
5 1 3 2 4 1 1 2 3 4

Sample Output 3

3

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
 
/* REPmacro */
#define REP(i,n) for(int i=0,i##_len=(n);i<i##_len;++i)
#define REPR(i,n) for(int i=((int)(n)-1);i>=0;--i)
#define FOR(i,m,n) for(int i=(int)(m);i<(int)(n);++i)
 
/*alias*/
#define ALL(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define F first
#define S second
#define PB push_back
#define EB emplace_back
#define MP make_pair 
 
#define y0 y3487465
#define y1 y8687969
#define j0 j1347829
#define j1 j234892
#define next asdnext
#define prev asdprev
#define itn int
 
/*using*/
using ll = long long;
using ull = unsigned long long;
using unsi = unsigned;
using vi = vector<int>;
using vvi = vector<vi>;
using vvvi = vector<vvi>;
using pii = pair<int, int>;
using db = double;
using plex = complex<double>;
using vs = vector<string>;
 
/*template*/
template<class T>inline bool amax(T &a, const T &b) { if (a < b) { a = b; return 1; } return 0; }
template<class T>inline bool amin(T &a, const T &b) { if (b < a) { a = b; return 1; } return 0; }
 
/* func */
 
/*struct*/
struct aaa {
	aaa() {
		cin.tie(0); ios::sync_with_stdio(0); cout << fixed << setprecision(20);
	};
}aaaaaaa;
 
/* const */
const int INF = 1001001001;
const ll LINF = 1001001001001001001ll;
const int MOD = 1e9 + 7; //10^9
const db EPS = 1e-9;
const int dx[] = { 1, 1, 0, -1, -1, -1, 0, 1 }, dy[] = { 0, 1, 1, 1, 0, -1, -1, -1 };
//Counterclockwise from the 9o'clock direction
//Take the x-axis downward and the y-axis right
 
//#define int ll
#define int long long
 
 array<int,200010> a;
signed main()
{
  int n, k;
  cin>>n>>k;
  for(int i = 0; i < n; ++i){
    int t;
    cin>>t;
    ++a[t];
  }
  
//  REP(i,n)cout<<a[i];

  sort(&a[0],&a[n+1]); //0~nとその次を対象にsort
/*
  int* itr2 = &a[0]+n+1;
  cout << *itr2 <<endl;
  auto itr=a.end();
  cout << *itr<<endl;
  */
  int ans = 0;
  for(int i = 0; i <= n-k; ++i)
    ans += a[i];
    cout<<ans;
 
  return 0;
}