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 cnt[205];
char s[100005], ans[100005];
int n, m, cntl, nbr[205];
int main() {
  scanf("%d%d%s", &n, &m, s);
  for (int i = 0; i < n; i++) {
    cnt[s[i]] = 1;
  }
  for (int i = 'a'; i <= 'z'; i++)
    if (cnt[i]) {
      cnt[i] = ++cntl;
      nbr[cntl] = i;
    }
  if (m > n) {
    for (int i = 0; i < n; i++) printf("%c", s[i]);
    for (int i = n; i < m; i++) printf("%c", nbr[1]);
    return 0;
  }
  int now = m - 1;
  for (now; now >= 0; now--) {
    if (s[now] == nbr[cntl])
      ans[now] = nbr[1];
    else {
      ans[now] = nbr[cnt[s[now]] + 1];
      break;
    }
  }
  for (int i = 0; i < now; i++) ans[i] = s[i];
  for (int i = 0; i < m; i++) printf("%c", ans[i]);
  return 0;
}
