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 MaxN = 1e6;
char s[200000];
int b[100];
char a[100];
int n, k, tnt, flag;
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> k;
  cin >> s;
  for (int i = 0; i < n; i++) {
    b[s[i] - 'a']++;
  }
  for (int i = 0; i < 26; i++) {
    if (b[i] != 0) {
      a[tnt++] = 'a' + i;
    }
  }
  tnt--;
  if (k > n) {
    for (int i = n; i < k; i++) {
      s[i] = a[0];
    }
    cout << s << endl;
    return 0;
  }
  s[k] = '\0';
  for (int i = k - 1; i >= 0; i--) {
    if (s[i] == a[tnt]) {
      s[i] = a[0];
    } else {
      for (int j = 0; j <= tnt; j++) {
        if (s[i] < a[j]) {
          flag = j;
          break;
        }
      }
      s[i] = a[flag];
      break;
    }
  }
  cout << s << endl;
  return 0;
}
