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
#ifndef BZ
#pragma GCC optimize "-O3"
#endif
#include <bits/stdc++.h>

#define ALL(v) (v).begin(), (v).end()
#define rep(i, l, r) for (int i = (l); i < (r); ++i)

using ll = long long;
using ld = long double;
using ull = unsigned long long;

using namespace std;


/*
ll pw(ll a, ll b) {
	ll ans = 1; while (b) {
		while (!(b & 1)) b >>= 1, a = (a * a) % MOD;
		ans = (ans * a) % MOD, --b;
	} return ans;
}
*/

string s;

int n;
int k;

int main() {
	ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
	cout.setf(ios::fixed), cout.precision(20);
	cin >> n >> k;
	cin >> s;
	string s2 = s;
	reverse(ALL(s2));
	string ans = s;
	s = s + s2;
	for (int i = 1; i + n <= s.size(); ++i)
		ans = min(ans, s.substr(i, n));
	--k;
	int cnt = 1;
	while (cnt < ans.size() && ans[0] == ans[cnt])
		++cnt;
	k = min(k, 20);
	int tmp = cnt;
	for (int i = 0; i < k; ++i)
		cnt = min(n, cnt * 2);
	for (int i = 0; i < cnt; ++i)
		cout << ans[0];
	for (int i = 0; i < n - cnt; ++i)
		cout << ans[tmp + i];
	cout << "\n";
	return 0;
}


