You are given a string ss of length nn, which consists only of the first kk letters of the Latin alphabet. All letters in string ss are uppercase.

A subsequence of string ss is a string that can be derived from ss by deleting some of its symbols without changing the order of the remaining symbols. For example, "ADE" and "BD" are subsequences of "ABCDE", but "DEA" is not.

A subsequence of ss called good if the number of occurences of each of the first kk letters of the alphabet is the same.

Find the length of the longest good subsequence of ss.

Input

The first line of the input contains integers nn (1≤n≤1051≤n≤105) and kk (1≤k≤261≤k≤26).

The second line of the input contains the string ss of length nn. String ss only contains uppercase letters from 'A' to the kk-th letter of Latin alphabet.

Output

Print the only integer — the length of the longest good subsequence of string ss.

Examples

Input

9 3
ACAABCCAB

Output

6

Input

9 4
ABCABCABC

Output

0

Note

In the first example, "ACBCAB" ("ACAABCCAB") is one of the subsequences that has the same frequency of 'A', 'B' and 'C'. Subsequence "CAB" also has the same frequency of these letters, but doesn't have the maximum possible length.

In the second example, none of the subsequences can have 'D', hence the answer is 00.

 

题意：给一个字符串，抽出一些字符使得每个字符都出现相同的次数，问最长能有多长。

思路：找到出现次数最小的字符，记录其出现次数minn，输出minn*n

#include<stdio.h>
#include<string.h>
#include<string>
#include<algorithm>
using namespace std;
#define INF 0x3f3f3f3;
int cal[26];
char ch[100010];
int index(char ch)
{
	return ch - 'A';
}
int main()
{
	int m, n;
	while (~scanf("%d%d", &m, &n))
	{
		int minn = INF;
		memset(cal, 0, sizeof(cal));
		scanf("%s", ch);
		for (int i = 0; i < m; i++)
		{
			cal[index(ch[i])]++;
		}
		for (int i = 0; i < n; i++)
		{
			minn = min(minn, cal[i]);
		}
		printf("%d\n", minn*n);
	}
	return 0;
}