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<bits/stdc++.h>
using namespace std;
const int N=1e4+5;
char c[N],s[N],t[N];
int n,k,Debug;
inline bool check(char *c,char *t){
	for(register int i=0;i<n;i++)
		if(c[i]>t[i]){
			if(Debug)	cout<<i<<' '<<c[i]<<' '<<t[i]<<endl;
			return false;
		}
		else	if(c[i]<t[i])	return true;
	return false;
}
inline bool run(){
	for(register int i=n-1;~i;i--)
		if(s[i]!=s[n-1])	return true;
	return false;
}
inline void com(char *c,char *t){
	if(check(c,t))	for(register int i=0;i<n;i++)	t[i]=c[i];
}
int main(){
	scanf("%d%d%s",&n,&k,s),k--;
	for(register int i=0;i<n;i++)	s[2*n-1-i]=s[i];
	while(k--&&run()){
		for(register int i=0;i<n;i++)	t[i]='z';
		for(register int i=0;i<=n;i++){
			for(register int j=i,k=n-1;j<i+n;j++)
				c[k--]=s[j];
			com(c,t);
		}
		reverse(t,t+n);
		for(register int i=0;i<n;i++)	s[2*n-1-i]=s[i]=t[i];
	}
	for(register int i=0;i<n;i++)	t[i]='z';
	for(register int i=0;i<=n;i++){
		for(register int j=i,k=0;j<i+n;j++) c[k++]=s[j];
		com(c,t);
	}
	cout<<t;
	return 0;
}
