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() {
  int n, k;
  cin >> n >> k;
  string s;
  cin >> s;
  int cnt[26];
  for (int i = 0; i < 26; i++) {
    cnt[i] = 0;
  }
  int len = s.size();
  for (int i = 0; i < len; i++) {
    cnt[s[i] - 97]++;
  }
  string ans = "";
  if (n < k) {
    ans = s;
    for (int i = 0; i < 26; i++) {
      if (cnt[i] > 0) {
        for (int j = 0; j < k - n; j++) {
          ans += (char)(i + 97);
        }
        cout << ans;
        return 0;
      }
    }
  } else {
    int Max = 0, Min = -1;
    for (int i = 0; i < 26; i++) {
      if (cnt[i] > 0) {
        Max = i;
      }
      if (cnt[i] > 0 && Min == -1) {
        Min = i;
      }
    }
    for (int i = k - 1; i >= 0; i--) {
      if (s[i] - 97 < Max) {
        ans = s.substr(0, i);
        for (int j = s[i] - 97 + 1; j < 26; j++) {
          if (cnt[j] > 0) {
            ans += (char)(j + 97);
            for (int l = i + 1; l < k; l++) {
              ans += (char)(Min + 97);
            }
            cout << ans;
            return 0;
          }
        }
      }
    }
  }
}
