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>
using namespace std;
int main() {
  int n, k, c[26], i, j, min_char, max_char;
  char in[100005];
  while (scanf("%d%d%s", &n, &k, in) != EOF) {
    memset(c, 0, sizeof(int) * 26);
    min_char = 26;
    max_char = -1;
    for (i = 0; i < strlen(in); i++) {
      c[in[i] - 'a']++;
      min_char = min(min_char, in[i] - 'a');
      max_char = max(max_char, in[i] - 'a');
    }
    if (n < k) {
      printf("%s", in);
      for (i = n; i < k; i++) printf("%c", min_char + 'a');
      printf("\n");
    } else {
      for (i = k - 1; i >= 0; i--) {
        if (in[i] - 'a' == max_char) {
          in[i] = min_char + 'a';
        } else {
          for (j = in[i] - 'a' + 1; j < 26; j++) {
            if (c[j] > 0) break;
          }
          in[i] = j + 'a';
          break;
        }
      }
      in[k] = 0;
      printf("%s\n", in);
    }
  }
  return 0;
}
