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;
char s[100004], ans[100004];
int n, k, cnt[26];
int main() {
  scanf("%d%d", &n, &k);
  scanf("%s", s + 1);
  for (int i = 1; i <= n; i++) cnt[s[i] - 'a']++;
  if (k > n) {
    for (int i = 1; i <= n; i++) printf("%c", s[i]);
    for (int i = 0; i < 26; i++) {
      if (cnt[i]) {
        for (int j = 1; j <= k - n; j++) printf("%c", i + 'a');
        puts("");
        return 0;
      }
    }
  }
  bool ok = 0;
  int st = 0;
  for (int i = 0; i < 26; i++)
    if (cnt[i]) {
      st = i;
      break;
    }
  for (int i = k; i >= 1; i--) {
    int lad = -1;
    if (ok) {
      ans[i] = s[i];
      continue;
    }
    for (int j = int(s[i] - 'a') + 1; j < 26; j++)
      if (cnt[j]) {
        lad = j;
        break;
      }
    if (lad == -1)
      lad = st;
    else
      ok = 1;
    ans[i] = lad + 'a';
  }
  for (int i = 1; i <= k; i++) printf("%c", ans[i]);
  puts("");
  return 0;
}
