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 main(){
	int N,K;
	string S;
	cin >> N >> K >> S;
	
	char t='z';
	for(int i=0;i<N;i++)t=min(t,S[i]);
	
	string R=S;
	reverse(R.begin(),R.end());
	S+=R;
	
	int m=1,c=0;
	for(int i=0;i<2*N;i++){
		if(S[i]==t)c++;
		else{
			m=max(m,c);
			c=0;
		}
	}
	
	vector<int> A;
	c=0;
	for(int i=0;i<2*N;i++){
		if(S[i]==t){
			c++;
			if(c==m)A.push_back(i);
		}else c=0;
	}
	
	if(K>20)K=20;
	if(m*pow(2,K-1)>=N){
		string ans="";
		for(int i=0;i<N;i++)ans+=t;
		cout << ans << endl;
		return 0;
	}
	
	vector<string> ans;
	
	if(K==1){
		for(auto a:A){
			if(a-m+N<2*N)ans.push_back(S.substr(a-m+1,N));
		}
	}
	else{
		for(auto a:A){
			if(a>=N-1){
				string b=S.substr(a-N+1,N);
				string br=b;
				reverse(br.begin(),br.end());
				b+=br;
				for(int k=2;k<K;k++){
					b=b.substr(m*pow(2,k-2),N);
					br=b;
					reverse(br.begin(),br.end());
					b+=br;
				}
				b=b.substr(N-m*pow(2,K-2),N);
				ans.push_back(b);
			}
		}
	}
	
	sort(ans.begin(),ans.end());
	cout << ans[0] << endl;
	return 0;
}