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;
int n,k;char c[5010],ans[5010],s[10010],g;
void chk(char *s){
	for(int i=0;i<n;i++)if(s[i]!=ans[i]){
		if(s[i]>ans[i])return;
		for(int j=0;j<n;j++)ans[j]=s[j];
		return;
	}
}
void solve(char *s,int t){
	char e[10010];
	for(int i=0;i<n;i++)e[i]=s[i];
	for(int i=n;i<n*2;i++)e[i]=s[n+n-i-1];
	if(t==1){
		for(int i=n;i>=0;i--)if(e[i-1]!=g||i==0){
			return chk(e+i);
		}
	}
	else{
		for(int i=0;i<=n;i++)if(e[i+n]!=g&&e[i+n-1]==g){
			solve(e+i,t-1);
			return;
		}
	}
}
int main(){
	cin>>n>>k;scanf("%s",c);
	g=*min_element(c,c+n);
	if(k>15){
		for(int i=1;i<=n;i++)putchar(g);
		return 0;
	}
	for(int i=0;i<n;i++)ans[i]=c[i];
	for(int i=0;i<n;i++)s[i]=c[i];
	for(int i=n;i<n*2;i++)s[i]=c[n+n-i-1];
	for(int i=0;i<=n;i++)chk(s+i);
	if(k>1)for(int i=0;i<=n;i++)solve(s+i,k-1);
	for(int i=0;i<n;i++)putchar(ans[i]);
	return 0;
}
