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 sum[200005];
int main() {
  ios::sync_with_stdio(0);
  int t = 1;
  while (t--) {
    int n, k;
    cin >> n >> k;
    string s, t = "";
    cin >> s;
    map<char, int> make_pair;
    for (int i = 0; i < n; i++) make_pair[s[i]] = 1;
    for (int i = 0; i < min(k, n); i++) t += s[i];
    char mn;
    for (char i = 'a'; i <= 'z'; i++) {
      if (make_pair[i] == 1) {
        mn = i;
        break;
      }
    }
    if (k > n) {
      for (int i = n; i < k; i++) s += mn;
      cout << s << endl;
      return 0;
    }
    for (int i = k - 1; i >= 0; i--) {
      char ch = s[i];
      bool f = 0;
      for (char j = char(ch + 1); j <= 'z'; j++) {
        if (make_pair[j] == 1) {
          t[i] = j;
          f = 1;
          break;
        }
      }
      if (f == 1)
        break;
      else
        t[i] = mn;
    }
    cout << t << endl;
  }
  return 0;
}
