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 MAX = 100010;
int main() {
  int n, k, idx;
  scanf("%d %d", &n, &k);
  char given[MAX], target[MAX];
  scanf("%s", given);
  bool alpha[28];
  memset(alpha, false, sizeof alpha);
  for (int i = 0; given[i]; i++) alpha[given[i] - 'a'] = true;
  char ch;
  for (int i = 0; i < 26; i++)
    if (alpha[i]) {
      ch = i + 'a';
      break;
    }
  if (k <= n) {
    bool hoise = false;
    for (int i = k - 1; i >= 0; i--) {
      for (int j = given[i] - 'a' + 1; j < 26; j++) {
        if (alpha[j]) {
          target[i] = j + 'a';
          idx = i;
          hoise = true;
          break;
        }
      }
      if (hoise) break;
      target[i] = ch;
    }
    for (int i = 0; i < idx; i++) target[i] = given[i];
    target[k] = '\0';
    puts(target);
    return 0;
  }
  for (int i = 0; given[i]; i++) target[i] = given[i];
  for (int i = 0; i < 26; i++)
    if (alpha[i]) {
      idx = i;
      break;
    }
  for (int i = n; i < k; i++) target[i] = idx + 'a';
  target[k] = '\0';
  puts(target);
  return 0;
}
