Appleman and Card Game CodeForces - 462B （贪心）

Appleman and Card Game

CodeForces - 462B

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.

Example

Input

15 10
DZFDFZDFDDDDDDF

Output

82

Input

6 4
YJSNPI

Output

4

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.

贪心，直接储存下每种字母出现的次数，每次都选取出现次数最多的，次数的平方就是每次可以得到的钱

code：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
long long n,k;
char s[100010];
long long num[100];
bool cmp(long long a,long long b){
return a > b;
}
int main(){
scanf("%lld%lld",&n,&k);
scanf("%s",s);
int i;
memset(num,0,sizeof(num));
for(i = 0; i < n; i++){
num[s[i]-'A']++;
}
sort(num,num+26,cmp);
long long ans = 0;
for(i = 0; i < 26; i++){
if(num[i]<=k){
ans = ans+(num[i]*num[i]);
k -= num[i];
}
else{
ans = ans + k*k;
k =0;
}
if(!k)break;
}
printf("%lld\n",ans);
return 0;
}