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() {
  long long n, k, i, j, m, sum = 0, count = 0, ans = 0, x;
  cin >> n >> k;
  string s;
  cin >> s;
  char c = 'z';
  long long a[26] = {0};
  for (i = 0; i < n; i++) {
    if (s[i] < c) c = s[i];
    a[s[i] - 'a']++;
  }
  if (k > n) {
    cout << s;
    for (i = 0; i < (k - n); i++) cout << c;
  } else {
    m = 0;
    for (i = int(s[k - 1] - 'a') + 1; i < 26; i++) {
      if (a[i] > 0) {
        m = 1;
        break;
      }
    }
    if (m == 1) {
      j = i;
      for (i = 0; i < k - 1; i++) cout << s[i];
      cout << char('a' + j);
    } else {
      i = k - 1;
      while (s[k - 1] == s[i]) {
        i--;
      }
      m = 0;
      j = i;
      for (i = int(s[j] - 'a') + 1; i < 26; i++) {
        if (a[i] > 0) {
          m = 1;
          break;
        }
      }
      for (x = 0; x < k; x++) {
        if (x < j)
          cout << s[x];
        else if (x == j)
          cout << char('a' + i);
        else
          cout << c;
      }
    }
  }
  return 0;
}
