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;

const int maxn = 5005;

int n, k, cnt;
string s, ans[maxn];

int main()
{
	cin >> n >> k >> s;

	char minc = s[0]; 
	for (int i = 1; i < n; ++i) minc = min(minc, s[i]);

	int mxl = 0;
	for (int i = 0, len = 0; i < n; ++i)
		if (s[i] == minc) ++len;
		else mxl = max(mxl, len), len = 0;
	if (k >= 14) return cout << string(n, minc), 0;

	string tmp = s; reverse(tmp.begin(), tmp.end());
	s = s + tmp;
	for (int i = 0, len = 0; i < n; ++i) {
		if (s[i] == minc) ++len;
		else mxl = max(mxl, len), len = 0;
		if (i == n - 1) {
			if (s[i] != minc) break;
			++cnt;
			int L = len << k;
            if (L >= n) return cout << string(n, minc), 0;
			ans[cnt] = s.substr(n - len - (n - L), n - L); reverse(ans[cnt].begin(), ans[cnt].end());
			ans[cnt] = string(L, minc) + ans[cnt];
		} else if (len == mxl) {
			++cnt;
			int L = len << (k - 1);
			ans[cnt] = string(L, minc) + s.substr(i + 1, n - L);
		}
	}

	sort(ans + 1, ans + cnt + 1);
	cout << ans[1].substr(0, n) << endl;
	
	return 0;
}
