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;
bool exists[26];
int hi, lo;
string get_best(string s, int pos, int final_len) {
  int char_at_pos = s[pos] - 'a';
  if (char_at_pos == hi) {
    return "";
  }
  int next_best = -1;
  for (int i = char_at_pos + 1; i < 26; ++i) {
    if (exists[i] == true) {
      next_best = i;
      break;
    }
  }
  string res = s.substr(0, pos);
  res += 'a' + next_best;
  res.resize(final_len, 'a' + lo);
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  int n, k;
  cin >> n >> k;
  string s;
  cin >> s;
  for (int i = 0; i < n; ++i) exists[s[i] - 'a'] = true;
  hi = 0;
  lo = 26;
  for (int i = 0; i < 26; ++i) {
    if (exists[i]) {
      hi = max(i, hi);
      lo = min(i, lo);
    }
  }
  s.resize(k, 'a' + lo);
  if (k > n) {
    cout << s << endl;
    return 0;
  }
  for (int i = k - 1; i >= 0; --i) {
    string res = get_best(s, i, k);
    if (!res.empty()) {
      cout << res << endl;
      return 0;
    }
  }
  return 0;
}
