NUC_TeamTEST_B（贪心）

B - B

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status

Description

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.

Sample Input

Input

15 10
DZFDFZDFDDDDDDF

Output

82

Input

6 4
YJSNPI

Output

4

Hint

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.

 

 中间超int啊，不难的贪心。CodeForces，DIV2的A题

?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm>   using namespace std;   const int max_size = 100000+10;   int main() {     int n, k;     char str[max_size];     long long  arry[26];     while(scanf("%d %d", &n, &k) == 2)     {         getchar();         memset(arry, 0, sizeof(arry));         scanf("%s", str);         int len = strlen(str);         int i;         for(i = 0; i < len; ++i)         {             arry[str[i]-'A']++;         }         i = 25;         sort(arry, arry+26);         //printf("%d\n", arry[25]);         long long sum = 0;         long long now = k;         do{             if(now > arry[i])             {                 sum += arry[i] * arry[i];                 now -= arry[i];                 //printf("%d\n", sum);                 i--;             }else if(now <= arry[i]){                 sum += now*now;          ///这个比较坑，没考虑到超int                 break;             }         }while(now != 0);         cout << sum << endl;     }     return 0; }