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, cnt[30] = {0}, i, j;
  char s[100005];
  scanf("%d%d%s", &n, &k, &s);
  for (i = 0; i < n; i++) cnt[s[i] - 'a' + 1] = 1;
  if (n < k) {
    for (i = 1; i <= 26; i++)
      if (cnt[i] == 1) break;
    printf("%s", s);
    for (j = 0; j < k - n; j++) printf("%c", i + 'a' - 1);
  } else {
    int flag = 0, y;
    for (i = k - 1; i >= 0; i--) {
      for (j = s[i] - 'a' + 2; j <= 26; j++)
        if (cnt[j] != 0) {
          flag = 1, y = j;
          break;
        }
      if (flag) break;
    }
    s[i] = '\0';
    printf("%s", s);
    printf("%c", y + 'a' - 1);
    for (j = 1; j <= 26; j++)
      if (cnt[j] == 1) break;
    for (int x = 1; x < k - i; x++) printf("%c", j + 'a' - 1);
  }
  return 0;
}
