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() {
  ios_base::sync_with_stdio(false);
  int n, k;
  string s;
  cin >> n >> k >> s;
  set<char> chars;
  char last = 'a', first = 'z';
  for (int i = 0; i < s.length(); i++) {
    chars.insert(s[i]);
    last = max(last, s[i]);
    first = min(first, s[i]);
  }
  string res = "";
  if (k > s.length()) {
    res = s;
    k -= s.length();
    while (k) {
      res += first;
      k--;
    }
  } else {
    for (int i = 0; i < k; i++) res += s[i];
    for (int j = res.length() - 1; j >= 0; j--) {
      if (res[j] == last) {
        res[j] = first;
      } else {
        res[j] = *chars.upper_bound(res[j]);
        break;
      }
    }
  }
  cout << res << endl;
  return 0;
}
