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 N = 5103;

int n, m;
int maxl, tot;
char s[N], minc;
char res[N][N];

int main() {
  scanf("%d%d", &n, &m);
  scanf("%s", s + 1);
  minc = 'z';
  for (int i = 1; i <= n; ++i) minc = min(minc, s[i]);
  if (m >= 14) {
    for (int i = 1; i <= n; ++i) putchar(minc);
    return !puts("");
  }
  for (int i = 1, len = 0; i <= n; ++i) {
    if (s[i] != minc) len = 0;
    else len++;
    if (len > maxl) maxl = len;
  }
  for (int i = 1, len = 0; i <= n; ++i) {
    if (s[i] != minc) len = 0;
    else len++;
    if (len == maxl) {
      ++tot;
      int have = min(n, len * (1 << (m - 1)));
      for (int j = 1; j <= have; ++j) res[tot][j] = minc;
      for (int j = have + 1, k = i + 1; j <= n; ++j, ++k) {
        if (k <= n) res[tot][j] = s[k];
        else res[tot][j] = s[2 * n - k + 1];
      }
    }
    if (i == n) {
      ++tot;
      int have = min(n, len * (1 << m));
      for (int j = 1; j <= have; ++j) res[tot][j] = minc;
      for (int j = have + 1, k = n - len; j <= n; ++j, --k)
        res[tot][j] = s[k];
    }
  }
  int best = 1;
  for (int i = 2; i <= tot; ++i) {
    if (strcmp(res[i] + 1, res[best] + 1) < 0)
      best = i;
  }
  puts(res[best] + 1);
}
