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 N = 100010;
int main() {
  int n, k;
  string s;
  cin >> n >> k >> s;
  bool vis[30] = {false};
  for (char c : s) vis[c - 'a'] = true;
  if (k > n) {
    for (int i = 0; i < 26; i++) {
      if (vis[i]) {
        while (n < k) {
          s += char('a' + i);
          n++;
        }
        cout << s;
        return 0;
      }
    }
  }
  for (int i = k - 1; i >= 0; i--) {
    int l = int(s[i] - 'a') + 1;
    while (l < 26) {
      if (vis[l]) {
        s[i] = l + 'a';
        char c = 'a';
        for (int j = 0; j < 26; j++)
          if (vis[j]) {
            c = 'a' + j;
            break;
          }
        i++;
        while (i < k) s[i++] = c;
        cout << s.substr(0, k);
        return 0;
      }
      l++;
    }
  }
  return 0;
}
