【Codeforces Round 263 (Div 2)B】【水题贪心】Appleman and Card Game 选字符，每个字符个数的平方

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Appleman has n cards. Each card has an uppercase letter written on it. Toastman must choose k cards from Appleman's cards. Then Appleman should give Toastman some coins depending on the chosen cards. Formally, for each Toastman's card i you should calculate how much Toastman's cards have the letter equal to letter on ith, then sum up all these quantities, such a number of coins Appleman should give to Toastman.

Given the description of Appleman's cards. What is the maximum number of coins Toastman can get?

Input

The first line contains two integers n and k (1 ≤ k ≤ n ≤ 10⁵). The next line contains n uppercase letters without spaces — the i-th letter describes the i-th card of the Appleman.

Output

Print a single integer – the answer to the problem.

Examples

input

15 10
DZFDFZDFDDDDDDF

output

input

6 4
YJSNPI

output

Note

In the first test example Toastman can choose nine cards with letter D and one additional card with any letter. For each card with D he will get 9 coins and for the additional card he will get 1 coin.

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); }
#define MS(x,y) memset(x,y,sizeof(x))
#define MC(x,y) memcpy(x,y,sizeof(x))
#define MP(x,y) make_pair(x,y)
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b>a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b<a)a = b; }
const int N = 0, M = 0, Z = 1e9 + 7, ms63 = 0x3f3f3f3f;
int n, k;
int cnt[26];
int main()
{
	while (~scanf("%d%d", &n,&k))
	{
		for (int i = 1; i <= n; ++i)
		{
			char ch; scanf(" %c", &ch);
			ch -= 'A';
			++cnt[ch];
		}
		LL ans = 0;
		sort(cnt, cnt + 26);
		for (int i = 25; i >= 0; --i)
		{
			LL take = min(k, cnt[i]);
			k -= take;
			ans += take*take;
		}
		printf("%lld\n", ans);
	}
	return 0;
}
/*
【题意】
给你一个字符串，长度为n（1e5）
我们选其中的k个，使得∑每个字符i（i∈[1,k]）*该字符个数尽可能大

【类型】
水题 贪心

【分析】
直接贪心选择数量最多的字符即可

【时间复杂度&&优化】
O(n)
*/