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=10005;
char s[N];int p[N];
signed main(){
	int n,K;
	long long res=1;
	cin>>n>>K;
	for(int i=2;i<=K;++i){
		res*=2;
		if(res>=n){
			res=n; break;
		}
	}
	cin>>(s+1);
	for(int i=n+1;i<=n+n;++i)
		s[i]=s[i-n];
	reverse(s+n+1,s+n+n+1);
	char ch=*min_element(s+1,s+n+1);
	int cnt=(s[1]==ch);
	for(int i=2;i<=n+n;++i)
		if(s[i]==ch && s[i]==s[i-1]) ++cnt;
		else p[i-1]=cnt,cnt=(s[i]==ch);
	p[n+n]=cnt;
	int mx=*max_element(p+n+1,p+n+n+1);
	res*=mx;
	if(res>=n){
		for(int i=1;i<=n;++i)
			cout<<ch;
		return 0;
	}
	string S="",tmp;
	for(int i=n+1;i<=n+n;++i)
		if(p[i]==mx){
			tmp="";
			for(int j=1;j<=res;++j)
				tmp+=ch;
			for(int j=i-p[i];j>i-p[i]-(n-res);--j)
				tmp+=s[j];
			if(S=="") S=tmp;
			else S=min(S,tmp);
		}
	cout<<S;
	return 0;
}