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;
map<int, int> m;
pair<bool, char> check(int a, string s) {
  int flag = 0;
  char x = s[a], e;
  for (long long int i = 1 + (x - 'a'); i < 26; i++) {
    if (m.count(i)) {
      return {true, char(i + 'a')};
    }
  }
  return {false, 'f'};
}
int main() {
  ios_base ::sync_with_stdio(false), cin.tie(NULL), cout.tie(0);
  int a, b;
  cin >> a >> b;
  string s;
  cin >> s;
  for (long long int i = 0; i < s.size(); i++) {
    m[s[i] - 'a']++;
  }
  if (a < b) {
    cout << s;
    sort(s.begin(), s.end());
    for (long long int i = 0; i < b - a; i++) cout << s[0];
  } else {
    int k = 0;
    char e;
    for (long long int i = b - 1; i >= 0; i--) {
      pair<bool, char> p = check(i, s);
      if (p.first) {
        e = p.second;
        k = i;
        break;
      }
    }
    for (long long int i = 0; i < k; i++) cout << s[i];
    cout << e;
    sort(s.begin(), s.end());
    if (b - k - 1 > 0) {
      for (long long int i = 0; i < b - k - 1; i++) cout << s[0];
    }
  }
  return 0;
}
