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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, k;
  string second;
  cin >> n >> k;
  cin >> second;
  int len = second.length();
  int a[len];
  int present[26];
  for (int i = 0; i < 26; ++i) {
    present[i] = 0;
  }
  for (int i = 0; i < len; ++i) {
    a[i] = second[i];
    present[second[i] - 'a']++;
  }
  char ch = *min_element(a, a + len);
  if (n < k) {
    cout << second;
    for (int i = 1; i <= k - n; ++i) {
      cout << ch;
    }
    cout << endl;
  } else {
    string s1 = second;
    int stop = -1;
    for (int i = k - 1; i >= 0; i--) {
      for (int j = second[i] - 'a' + 1; j < 26; j++) {
        if (present[j]) {
          s1[i] = j + 'a';
          stop = i;
          break;
        }
      }
      if (stop > 0) {
        break;
      }
    }
    for (int j = 0; j <= min(stop, k - 1); ++j) {
      cout << s1[j];
    }
    for (int j = min(stop, k - 1) + 1; j < k; ++j) {
      cout << ch;
    }
    cout << endl;
  }
  return 0;
}
