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);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  string s, s1;
  cin >> s;
  map<char, int> mp;
  int i, j;
  s1 = s;
  for (i = 0; i < s.length(); i++) mp[s[i]]++;
  if (n < k) {
    sort(s1.begin(), s1.end());
    cout << s;
    k = k - n;
    for (i = 0; i < k; i++) cout << s1[0];
    cout << "\n";
  } else {
    sort(s1.begin(), s1.end());
    char ans[k + 1];
    int flag = 0;
    for (i = k - 1; i >= 0; i--) {
      char ch = s[i];
      if (flag == 1) {
        ans[i] = ch;
        continue;
      }
      for (char ch1 = ch + 1; ch1 <= 'z'; ch1++) {
        if (mp[ch1] > 0) {
          flag = 1;
          ans[i] = ch1;
          break;
        }
      }
      if (flag == 0) ans[i] = s1[0];
    }
    for (i = 0; i < k; i++) cout << ans[i];
    cout << "\n";
  }
  return 0;
}
