Takahashi has a string S of length N consisting of lowercase English letters. On this string, he will perform the following operation K times:

* Let T be the string obtained by reversing S, and U be the string obtained by concatenating S and T in this order.
* Let S' be some contiguous substring of U with length N, and replace S with S'.



Among the strings that can be the string S after the K operations, find the lexicographically smallest possible one.

Constraints

* 1 \leq N \leq 5000
* 1 \leq K \leq 10^9
* |S|=N
* S consists of lowercase English letters.

Input

Input is given from Standard Input in the following format:


N K
S


Output

Print the lexicographically smallest possible string that can be the string S after the K operations.

Examples

Input

5 1
bacba


Output

aabca


Input

10 2
bbaabbbaab


Output

aaaabbaabb
#include <iostream>
#include <cstdio>
#define MN 10010

char s[MN];
int l = -1, r = -1;

bool cmp(int a, int b, int c, int d)
{
	int len = b - a + 1;
	for(int i = 0; i < len; i++)
	{
		if(s[b - i] < s[d - i]) return 1;
		if(s[b - i] > s[d - i]) return 0;
	}
	return 0;
}

int main()
{
	int n, k; scanf("%d%d%s", &n, &k, s);
	for(int i = n; i < 2 * n; i++) s[i] = s[2 * n - i - 1];
	char Min = 'z'; int len = 0, Max = 0;
	for(int i = 0; i < 2 * n; i++)
	{
		if(s[i] < Min) Min = s[i], len = 1;
		else if(s[i] == Min) len++;
		else len = 0;
		Max = std::max(Max, len);
	}
	
	k--;
	
	if(k > 30 || ((1ll * Max) << k) >= n)
	{
		for(int i = 0; i < n; i++) printf("%c", Min);
		return 0;
	}
	
	len = 0; 
	
	for(int i = 0; i < 2 * n; i++)
	{
		if(s[i] == Min) len++;
		else len = 0;
		if(i >= n && len == Max)
		{
			int L = i - Max - (n - (Max << k)) + 1, R = i - Max;
			if(l == -1 || cmp(L, R, l, r)) l = L, r = R;
		}
	}
	
	for(int i = 1; i <= (Max << k); i++) printf("%c", Min);
	for(int i = r; i >= l; i--) printf("%c", s[i]);
}