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<algorithm>
#include<cstdio>
#include<cstring>
#include<string>
using namespace std;
string S,ans;
int main()
{
	int n,k;
	cin>>n>>k;
	cin>>S;
	char mnc='z'+1;
	for(int i=0;i<n;i++)
		mnc=min(mnc,S[i]);
	for(int i=n-1;i>=0;i--)
		S+=S[i];
	int len=0;
	int cur=0;
	for(int i=0;i<n*2;i++)
	{
		if(S[i]==mnc)cur++;
		else cur=0;
		len=max(len,cur);
	}
	int llen=len;
	for(int i=1;i<k;i++)
	{
		llen<<=1;
		if(llen>=n){
			for(int j=0;j<n;j++)
				printf("%c",mnc);
			printf("\n");
			return 0;
		}
	}
	for(int i=0;i<n-llen;i++)
		ans+='z'+1;
	cur=0;
	for(int i=0;i<n*2;i++)
	{
		if(S[i]==mnc)cur++;
		else cur=0;
		if(cur==len&&i>=n){
			string tmp=S.substr(i-n-len+llen+1,n-llen);
			for(int j=0;j<n-llen-j-1;j++)
				swap(tmp[j],tmp[n-llen-1-j]);
			ans=min(ans,tmp);
		}
	}
	for(int i=0;i<llen;i++)
		printf("%c",mnc);
	cout<<ans<<"\n";
	return 0;
}