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(0);
  cin.tie(0);
  ;
  int n, k;
  cin >> n >> k;
  string s;
  cin >> s;
  vector<char> letters;
  for (auto e : s) letters.push_back(e);
  sort((letters).begin(), (letters).end());
  letters.resize(
      distance(letters.begin(), unique((letters).begin(), (letters).end())));
  string t = s;
  if (k <= s.size())
    t.resize(k);
  else {
    int m = k - ((int)t.size());
    for (int i = 0; i < m; i++) t += letters[0];
    cout << t << endl;
    return 0;
  }
  int inv[200];
  int m = letters.size();
  for (int i = 0; i < m; i++) inv[letters[i]] = i;
  for (int i = t.size() - 1; i >= 0; i--) {
    int j = inv[t[i]];
    if (j + 1 < m) {
      t[i] = letters[j + 1];
      for (int l = i + 1; l < t.size(); l++) t[l] = letters[0];
      break;
    }
  }
  cout << t << endl;
  return 0;
}
