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;
const int N = 100005;
char s[N], t[N];
int n, m, vst[30], mn = 30, mx = -1;
int main() {
  scanf("%d%d", &n, &m);
  scanf("%s", s);
  int i, j;
  for (i = 0; i < n; i++) {
    j = s[i] - 'a';
    vst[j] = 1;
    mx = max(mx, j);
    mn = min(mn, j);
  }
  if (m > n) {
    for (i = 0; i < n; i++) printf("%c", s[i]);
    for (i = n; i < m; i++) printf("%c", char('a' + mn));
    return 0;
  }
  int t;
  for (i = m - 1; i >= 0; i--) {
    if (s[i] - 'a' != mx) {
      for (t = s[i] - 'a' + 1;; t++)
        if (vst[t]) break;
      for (j = 0; j < i; j++) printf("%c", s[j]);
      printf("%c", char('a' + t));
      for (j = i + 1; j < m; j++) printf("%c", char('a' + mn));
      return 0;
    }
  }
  return 0;
}
