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 <vector>
#include <cstdio>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;

const int N=1e4+5;
int n,k,tmp,Mx,len[N];
char str[N],ans[N],Mi='z';

bool flag=false;
inline void copy(int pos) {
	for(int i=1;i<=n;++i) ans[i]=str[pos+i-1];
}
inline void solve(int pos) {
	if(!flag) return flag=true,copy(pos),void();
	for(int i=1;i<=n;++i) 
		if(ans[i]<str[pos+i-1]) return void();
		else if(ans[i]>str[pos+i-1]) return copy(pos),void();
}

int main() {
	scanf("%d%d%s",&n,&k,str+1);
	for(int i=1;i<=n;++i) Mi=min(Mi,str[i]);
	for(int i=1;i<=n;++i) str[(n<<1)-i+1]=str[i];
	if(k>=15) {for(int i=1;i<=n;++i) putchar(Mi);return 0;}
	if(k==1) {
		for(int i=1;i<=n+1;++i) solve(i);
		return printf("%s\n",ans+1),0;
	}
	for(int i=1;i<=n+n;++i) if(str[i]==Mi) len[i]=len[i-1]+1;
	for(int i=1;i<=n+n;++i) Mx=max(Mx,len[i]);
	tmp=Mx<<k-1;
	if(tmp>=n) {for(int i=1;i<=n;++i) putchar(Mi);return 0;}
	for(int i=1;i<=n+Mx;++i) if(len[i]==Mx) solve(i-len[i]+1);
	for(int i=1;i<=tmp;++i) putchar(Mi);
	for(int i=1;i<=n-tmp;++i) putchar(ans[i+Mx]);
	return 0;
}