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 str[100010];
int vis[30];
int main() {
  int n, m;
  while (scanf("%d%d", &n, &m) != EOF) {
    scanf("%s", str);
    memset(vis, 0, sizeof(vis));
    for (int i = 0; str[i]; i++) vis[str[i] - 'a'] = 1;
    if (m > n) {
      printf("%s", str);
      int x;
      for (int i = 0; i < 26; i++)
        if (vis[i]) {
          x = i;
          break;
        }
      for (int i = n + 1; i <= m; i++) printf("%c", x + 'a');
    } else {
      int x;
      for (int i = 25; i >= 0; i--) {
        if (vis[i]) {
          x = i;
          break;
        }
      }
      for (int i = m - 1; i >= 0; i--) {
        if (str[i] != x + 'a') {
          for (int j = 0; j < i; j++) printf("%c", str[j]);
          for (int j = 0; j < 26; j++)
            if (vis[j] && j > str[i] - 'a') {
              printf("%c", j + 'a');
              break;
            }
          for (int j = 0; j < 26; j++) {
            if (vis[j]) {
              for (int k = i + 1; k < m; k++) printf("%c", j + 'a');
              break;
            }
          }
          break;
        }
      }
    }
    printf("\n");
  }
  return 0;
}
