And where the are the phone numbers?

You are given a string s consisting of lowercase English letters and an integer k. Find the lexicographically smallest string t of length k, such that its set of letters is a subset of the set of letters of s and s is lexicographically smaller than t.

It's guaranteed that the answer exists.

Note that the set of letters is a set, not a multiset. For example, the set of letters of abadaba is {a, b, d}.

String p is lexicographically smaller than string q, if p is a prefix of q, is not equal to q or there exists i, such that pi < qi and for all j < i it is satisfied that pj = qj. For example, abc is lexicographically smaller than abcd , abd is lexicographically smaller than abec, afa is not lexicographically smaller than ab and a is not lexicographically smaller than a.

Input

The first line of input contains two space separated integers n and k (1 ≤ n, k ≤ 100 000) — the length of s and the required length of t.

The second line of input contains the string s consisting of n lowercase English letters.

Output

Output the string t conforming to the requirements above.

It's guaranteed that the answer exists.

Examples

Input

3 3
abc


Output

aca


Input

3 2
abc


Output

ac


Input

3 3
ayy


Output

yaa


Input

2 3
ba


Output

baa

Note

In the first example the list of strings t of length 3, such that the set of letters of t is a subset of letters of s is as follows: aaa, aab, aac, aba, abb, abc, aca, acb, .... Among them, those are lexicographically greater than abc: aca, acb, .... Out of those the lexicographically smallest is aca.
#include <bits/stdc++.h>
int main() {
  char s[100005];
  int i, j, n, k;
  int len;
  int vis[30] = {0};
  int count = 0;
  scanf("%d %d", &n, &k);
  scanf("%s", s);
  for (i = 0; i < n; i++) {
    vis[s[i] - 'a' + 1] = 1;
  }
  int max = 0;
  int min = 30;
  for (i = 1; i <= 26; i++) {
    if (vis[i] == 1 && max < i) max = i;
    if (vis[i] == 1 && min > i) min = i;
  }
  for (i = 1; i <= 26; i++) {
    if (vis[i] == 1) count++;
  }
  if (n >= k) {
    for (i = k - 1; i >= 0; i--) {
      if (s[i] - 'a' + 1 < max) {
        for (j = 1; j <= 26; j++) {
          if (vis[j] == 1 && s[i] - 'a' + 1 < j) {
            s[i] = 'a' + j - 1;
            break;
          }
        }
        for (j = i + 1; j <= k - 1; j++) {
          s[j] = 'a' + min - 1;
        }
        break;
      }
    }
    for (i = 0; i < k; i++) {
      printf("%c", s[i]);
    }
    printf("\n");
  }
  if (n < k) {
    printf("%s", s);
    for (i = 1; i <= k - n; i++) {
      printf("%c", min + 'a' - 1);
    }
    printf("\n");
  }
  return 0;
}
