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)

/* -------------------------------------------------

Author :- Vaibhav Pal
College :- Northern India Engineering College, New Delhi
Coding Handle :- underdog_eagle
Title :-

----------------------------------------------------*/
#include <bits/stdc++.h>
#include <unordered_map>
#include <unordered_set>
using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> ii;
typedef pair<ll, ll> pll;
typedef vector<ii> vii;
typedef vector<pll> vpll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<vi> vvi;

#define FAST_IO ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define INF 1000000000000000000         
#define nl '\n'
#define sp ' '
#define pb push_back
#define mp make_pair
#define mt make_tuple
#define F first
#define S second
#define bg begin()
#define en end()
#define fo(i, a, b) for(int (i) = (a); (i) <= (b); (i)++)

inline ll gcd(ll a, ll b) { return b == 0 ? a : gcd(b, a % b); }
inline ll lcm(ll a, ll b) { return a * (b / gcd(a, b)); }

int main()
{
	FAST_IO
   /* #ifndef ONLINE_JUDGE
	freopen("in.txt", "r", stdin);
	freopen("out.txt", "w", stdout);
    #endif*/

	ll n, k;
	cin >> n >> k;
	map<ll, ll> m;
	map<ll, ll>::iterator it;
	fo(i, 1, n){
		ll x;
		cin >> x;
		it = m.find(x);
		if(it != m.en) it->second++;
		else m.insert(mp(x, 1));
	}
	
	vll v;
	for(it = m.bg; it != m.en; it++) v.pb(it->second);
	sort(v.bg, v.en);
	ll y = v.size();
	ll ans = 0;
	if(y > k)
		fo(i, 0, y - k - 1) ans += v[i];
	cout << ans;
	
	
	return 0;
}