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>
char s[100005];
int vis[300];
int flag[30];
int main() {
  int n, m;
  while (~scanf("%d%d", &n, &m)) {
    scanf("%s", s);
    memset(vis, 0, sizeof(vis));
    for (int i = 0; i < strlen(s); i++) vis[s[i]] = 1;
    int k = 0;
    for (int i = 50; i <= 200; i++)
      if (vis[i] == 1) flag[k++] = i;
    if (m > n) {
      int minn = 200;
      for (int i = 0; i < n; i++)
        if (s[i] < minn) minn = s[i];
      printf("%s", s);
      for (int i = n; i < m; i++) printf("%c", minn);
      printf("\n");
    } else {
      int fflag = 0;
      for (int i = m - 1; i >= 0; i--) {
        if (s[i] != flag[k - 1]) {
          for (int j = 0; j < k; j++) {
            if (s[i] == flag[j]) {
              s[i] = flag[j + 1];
              fflag = 1;
              break;
            }
          }
          if (fflag == 1) break;
        } else {
          s[i] = flag[0];
        }
      }
      for (int i = 0; i < m; i++) printf("%c", s[i]);
      printf("\n");
    }
  }
  return 0;
}
