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;
//#define cerr if (false) cerr
#define db(x) cerr << #x << "=" << x << endl
#define db2(x, y) cerr << #x << "=" << x << "," << #y << "=" << y << endl
#define db3(x, y, z) cerr << #x << "=" << x << "," << #y << "=" << y << "," << #z << "=" << z << endl
#define dbv(v) cerr << #v << "="; for (auto _x : v) cerr << _x << ", "; cerr << endl
#define dba(a, n) cerr << #a << "="; for (int _i = 0; _i < (n); ++_i) cerr << a[_i] << ", "; cerr << endl
typedef long long ll;
typedef long double ld;
char str[5005];
int main() {
	int n, k;
	scanf("%d%d%s", &n, &k, str);
	string s = str;
	string t(s.rbegin(), s.rend());
	s += t;
	string ans(n, 'z');
	for (int i = 0; i <= n; ++i) {
		t = s.substr(i, n);
		int c = 1;
		for (; c < n && t[n - c - 1] == t[n - 1]; ++c);
		for (int j = 1; c < n && j < k; ++j) {
			string tt(t.rbegin(), t.rend());
			t += tt;
			if (j == k - 1) t = t.substr(n - c, n);
			else t = t.substr(c, n);
			c = min(n, c * 2);
		}
		ans = min(ans, t);
	}
	printf("%s\n", ans.c_str());
}
