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 = 1e5 + 5;
int mp[26];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  string s;
  cin >> s;
  for (int i = 0; i < n; i++) {
    mp[s[i] - 'a']++;
  }
  int min1;
  for (int i = 0; i < 26; i++) {
    if (mp[i]) {
      min1 = i;
      break;
    }
  }
  if (n < k) {
    cout << s.substr(0, n);
    for (int i = 0; i < k - n; i++) {
      cout << char('a' + min1);
    }
    cout << '\n';
    return 0;
  }
  int g = -1;
  int wei = -1;
  for (int i = k - 1; i >= 0; i--) {
    if (g != -1) break;
    for (int j = 0; j < 26; j++) {
      int x = s[i] - 'a';
      if (mp[j] && j > x) {
        g = j;
        wei = i;
        break;
      }
    }
  }
  if (wei != 0) {
    cout << s.substr(0, wei);
  }
  cout << char('a' + g);
  for (int i = 0; i < k - (wei + 1); i++) {
    cout << char(min1 + 'a');
  }
  cout << '\n';
}
