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<cmath>
#include<vector>
using namespace std;
#define MAXN 10000
#define LL long long
#define Pr pair<int,int>
#define X first
#define Y second
int n,k;
char s[MAXN+5],Mn,Mx,ans[MAXN+5],tmp[MAXN+5];
LL read(){
    LL x=0,F=1;char c=getchar();
    while(c<'0'||c>'9'){if(c=='-')F=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*F;
}
void cal(char *s,int t){
	int np=n,len=0;
	while(np>=1&&s[np]==Mn)
	np--,len++;
	while(t>0&&len<n)len*=2,t--;
	if(len>=n){
		for(int i=1;i<=n;i++)
		ans[i]=Mn;
		return ;
	}
	for(int i=1;i<=len;i++)tmp[i]=Mn;
	for(int i=len+1;i<=n;i++)
	tmp[i]=s[np--];
	for(int i=1;i<=n;i++)
	if(tmp[i]<ans[i]){
		for(int j=1;j<=n;j++)
		ans[j]=tmp[j];
		return ;
	}else if(tmp[i]>ans[i])return ;
}
int main()
{
	n=read(),k=read();
	scanf("%s",s+1);
	Mn='z',Mx='a';
	for(int i=1;i<=n;i++){
		Mn=min(Mn,s[i]),Mx=max(Mx,s[i]);
		s[n*2-i+1]=s[i];ans[i]='z';
	}
	if(s[n]==Mn)cal(s,k);
	for(int i=n;i<=2*n;i++)
	if(s[i]==Mn)cal(s+i-n,k-1);
	printf("%s",ans+1);
}