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 <cstdio>
#include <algorithm>

const int MN = 10005;

int N, K;
char S[MN], U;
int A[MN], V, W;
char Ans[MN];

inline void check(int p) {
	int ok = 1;
	for (int i = 1; i <= W; ++i) {
		if (Ans[i] > S[p - i + 1]) break;
		if (Ans[i] < S[p - i + 1]) { ok = 0; break; }
	}
	if (!ok) return ;
	for (int i = 1; i <= W; ++i)
		Ans[i] = S[p - i + 1];
}

int main() {
	scanf("%d%d", &N, &K);
	scanf("%s", S + 1);
	for (int i = 1; i <= N; ++i) S[N + N - i + 1] = S[i];
	S[N + N + 1] = '\0';
	U = *std::min_element(S + 1, S + N + N + 1);
	for (int i = 1; i <= N + N; ++i)
		if (S[i] == U) A[i] = A[i - 1] + 1;
		else A[i] = 0;
	V = *std::max_element(A + 1, A + N + N + 1);
	if (K >= 14 || ((N - 1) >> (K - 1)) + 1 <= V) {
		for (int i = 1; i <= N; ++i) printf("%c", U);
		return 0;
	}
	Ans[1] = 'z' + 1;
	W = N - (V << (K - 1));
	for (int i = 1; i <= N; ++i)
		if (A[N + i] == V) check(N + i - V);
	for (int i = 1; i <= (V << (K - 1)); ++i) printf("%c", U);
	printf("%s\n", Ans + 1);
	return 0;
}