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;
  cin >> n >> k;
  string s;
  cin >> s;
  int set[26];
  int i;
  for (i = 0; i < 26; i++) set[i] = 0;
  for (i = 0; i < s.length(); i++) {
    set[s[i] - 'a'] = 1;
  }
  int least = 26;
  for (i = 0; i < 26; i++)
    if (set[i]) least = min(least, i);
  string ans;
  if (n < k) {
    ans = s;
    for (i = 1; i <= k - n; i++) {
      ans += string(1, least + 'a');
    }
  } else {
    for (i = 0; i < k; i++) ans += (char)s[i];
    for (i = k - 1; i >= 0; i--) {
      int num = s[i] - 'a';
      int exist = -1;
      int j;
      for (j = num + 1; j < 26; j++) {
        if (set[j]) {
          exist = j;
          break;
        }
      }
      if (exist >= 0) {
        ans[i] = (char)(j + 'a');
        break;
      } else {
        ans[i] = (char)(least + 'a');
      }
    }
  }
  cout << ans << endl;
  return 0;
}
