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 a[100002];
char b[26], c[100002];
int go(int now) {
  int chk = 0;
  for (int i = a[now] - 'a' + 1; i < 26; i++) {
    if (b[i]) {
      chk = 1;
      c[now] = i + 'a';
      break;
    }
  }
  if (chk) return now;
  return go(now - 1);
}
int main() {
  int n, k;
  scanf("%d %d", &n, &k);
  scanf("%s", a);
  int min = 300;
  for (int i = 0; i < n; i++) {
    if (!b[a[i] - 'a']) b[a[i] - 'a'] = 1;
    if (min > a[i] - 'a') min = a[i] - 'a';
  }
  strcpy(c, a);
  int now = go(k - 1);
  int len = strlen(a);
  len = now + 1 > len ? len : now + 1;
  for (int i = len; i < k; i++) {
    c[i] = min + 'a';
  }
  c[k] = 0;
  printf("%s\n", c);
}
