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 x[10005], mi;

int main() {
	scanf("%d%d%s", &n, &k, x);
	for (int i = 0; i < n; i++)
		x[n + i] = x[n - 1 - i];
	if (k == 1) {
		string ans = string(x, n);
		for (int i = 1; i <= n; i++)
			ans = min(ans, string(x + i, n));
		printf("%s\n", ans.c_str());
		return 0;
	}
	mi = x[0];
	for (int i = 1; i < n; i++)
		mi = min(mi, x[i]);
	string ans = string(x, n);
	for (int i = n + 1; i <= n + n; i++)
		if (x[i - 1] == mi && x[i] != mi) {
			int cnt = 0;
			for (int j = 1; i - j >= 0 && x[i - j] == mi; j++)
				cnt++;
			cnt = min(cnt, n);
			string cur = string(x + (n + n - (i - cnt)), n - cnt);
			if (cur.size() < ans.size() || (cur.size() == ans.size() && cur < ans))
				ans = cur;
		}
	int cnt = n - ans.size();
	for (int i = 1; i < k && cnt != n; i++)
		cnt = min(n, cnt * 2);
	for (int i = 0; i < cnt; i++)
		printf("%c", mi);
	printf("%s\n", string(ans.c_str(), n - cnt).c_str());
}
