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
//Love and Freedom.
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
#define ll long long
#define inf 20021225
#define N 10100
using namespace std;
int read()
{
	int s=0,t=1; char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')	t=-1; ch=getchar();}
	while(ch>='0' && ch<='9')	s=s*10+ch-'0',ch=getchar();
	return s*t;
}
char ch[N],ans[N]; int n,k;
int main()
{
	n=read(),k=read(); scanf("%s",ch+1); char mn=127;
	for(int i=1;i<=n;i++)	mn=min(mn,ch[i]);
	if(k>=16||(1<<k-1)>=n){for(int i=1;i<=n;i++)	printf("%c",mn); return 0;}
	for(int i=n+1;i<=2*n;i++)	ch[i]=ch[n-i+n+1],ans[i-n]=127;
	for(int i=1;i<=n+1;i++)
		for(int j=0;j<n;j++)	if(ans[j+1]!=ch[i+j])
		{
			if(ans[j+1]>ch[i+j])	for(int l=1;l<=n;l++)	ans[l]=ch[i+l-1];
			break;
		}
	int cnt=0;
	for(int i=1;i<=n;i++)	if(ans[i]==ans[1])	cnt++; else	break;
	int owo=min(n,cnt<<k-1);	for(int i=1;i<=owo;i++)	putchar(mn);
	for(int i=1;i<=n-owo;i++)	putchar(ans[i+cnt]); 
	return 0;
}