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
/*input
10 2
bbaabbbaab
*/

// bbaabbbaab

#include <bits/stdc++.h>
#define pb push_back
#define all(a) a.begin(), a.end()
#define sz(a) (int)a.size()
#define x first
#define y second
using namespace std;
using ll = long long;
using ld = long double;
using pii = pair<int, int>;
const int maxn = 2e5 + 100;
const int mod = 1e9 + 7;
int n, k;
string solve(string s)
{
	int it = n - 1;
	int cnt = 0;
	while(it >= 0 && s[it] == s[n - 1])
		++cnt, --it;
	for(int i = 0; i < k - 1; i++)
		cnt *= 2;
	if(cnt >= n)
		return string(n, s[n - 1]);
	else
	{
		string ret = string(cnt, s[n - 1]);
		while(cnt < n)
		{
			ret += s[it];
			--it;
			++cnt;
		}
		return ret;
	}
}
int main()
{
	ios_base::sync_with_stdio(false), cin.tie(0);
	cin >> n >> k;
	string s;
	cin >> s;

	char mn = 'z';
	for(int i = 0; i < n; i++)
		mn = min(mn, s[i]);

	if(k > 12)
		return cout << string(n, mn) << "\n", 0;

	string t = s;
	reverse(all(t));
	string st = s + t;

	string ans = s;
	for(int i = 0; i <= n; i++)
	{
		string ss = st.substr(i, n);
		ans = min(ans, solve(ss));
	}
	cout << ans << "\n";
	return 0;
}