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<iomanip>
#include<cstring>
#include<cmath>
#include<cstdio>
#include<queue>
#include<algorithm>
using namespace std;
int n,k;
int maxsum=0,sum=0,maxid;
char s[100005],ans[100005];
int main()
{
	scanf("%d%d",&n,&k);
	scanf("%s",s+1);
	for(int i=1;i<=n;i++) s[2*n-i+1]=s[i];
	for(int i=1;i<=n;i++) ans[i]=s[i];
	for(int i=2;i<=n;i++)
	{
		bool flag=1;
		for(int j=i;j<=n+i-1;j++)
		{
			if(ans[j-i+1]==s[j]) continue;
			if(ans[j-i+1]<s[j]) flag=0;
			break;
		}
		if(flag) for(int j=i;j<=n+i-1;j++) ans[j-i+1]=s[j];
	}
	for(int i=1;i<=n;i++)
	{
		if(ans[i]==ans[1]) maxsum=i;
		else break;
	}
	if(k>=16||1ll*maxsum*(1<<(k-1))>=n) for(int i=1;i<=n;i++) printf("%c",ans[1]);
	else
	{
		long long len=(1ll<<(k-1))*maxsum;
		for(int i=1;i<=len;i++) printf("%c",ans[1]);
		for(int i=1;i<=n-len;i++) printf("%c",ans[i+maxsum]);
	}
}