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;
 
typedef long long ll;
typedef pair<int, int> ii;

const int N = 3e5+100;

string ss[N];

string extend(string s){
	string t = s;
	reverse(t.begin(), t.end());
	s.insert(s.end(), t.begin(), t.end());
	return s;
}

signed main(){
	ios_base::sync_with_stdio(false);cin.tie(NULL);
	int n, k;cin >> n >> k;
	string s;cin >> s;
	char c = *min_element(s.begin(), s.end());
	if(k>14){
		for(int i=0 ; i<n ; i++)
			cout << c;
		cout << "\n";
		return 0;
	}
	if(k == 1){
		s = extend(s);
		for(int i=0 ; i<=n ; i++)
			ss[i] = s.substr(i, n);
		sort(ss, ss+n+1);
		cout << ss[0] << "\n";
		return 0;
	}
	k --;
	s = extend(s);
	for(int i=0 ; i<=n ; i++){
		ss[i] = s.substr(i, n);
		reverse(ss[i].begin(), ss[i].end());
	}
	sort(ss, ss+n+1);
	s = ss[0];
	reverse(s.begin(), s.end());
	int cnt=0;
	for(int i=s.size()-1 ; i>=0 ; i--)
		if(s[i] == s.back())
			cnt ++;
		else
			break;
	while(k){
		for(int i=0 ; i<cnt ; i++)
			s.push_back(s.back());
		if(cnt>n)
			break;
		k --;
		cnt *= 2;
	}
	s = s.substr(s.size()-n, n);
	reverse(s.begin(), s.end());
	cout << s << "\n";
 	return 0;
}














