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 <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define N 10005
typedef long long ll;
int n, k;
char Min, Max, ans[N], S[N];
void Check(char *S, int k)
{
	char T[N];
	int now = n, len = 0;
	while (now >= 1 && S[now] == Min)
		now--, len++;
	while (k != 0 && len < n)
		len *= 2, k--;
	if (len >= n)
	{
		for (int i = 1; i <= n; i++)
			ans[i] = Min;
		printf("%s\n", ans + 1);
		exit(0);
	}
	for (int i = 1; i <= len; i++)
		T[i] = Min;
	for (int i = len + 1; i <= n; i++)
		T[i] = S[now--];
	for (int i = 1; i <= n; i++)
		if (T[i] < ans[i])
		{
			for (int j = 1; j <= n; j++)
				ans[j] = T[j];
			return;
		}
		else
			if (T[i] > ans[i])
				return;
}
int main()
{
	scanf("%d%d", &n, &k);
	scanf("%s", S + 1);
	Min = 'z', Max = 'a';
	for (int i = 1; i <= n; i++)
	{
		Min = min(Min, S[i]);
		Max = max(Max, S[i]);
		S[2 * n - i + 1] = S[i];
		ans[i] = 'z';
	}
	if (Min == Max)
	{
		printf("%s\n", S + 1);
		return 0;
	}
	if (S[n] == Min)
		Check(S, k);
	for (int i = n; i <= 2 * n; i++)
		if (S[i] == Min)
			Check(S + i - n, k - 1);
	printf("%s\n", ans + 1);
}
