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() {
  int n, k;
  while (cin >> n >> k) {
    string str;
    cin >> str;
    set<char> s;
    for (int i = 0; i < str.size(); ++i) {
      s.insert(str[i]);
    }
    if (k > n) {
      for (int i = n; i < k; ++i) {
        str += *(s.begin());
      }
      cout << str << endl;
    } else {
      string ans;
      bool smaller = false;
      for (int i = k - 1; i >= 0; --i) {
        if (smaller) {
          ans = str[i] + ans;
        } else {
          auto it = s.upper_bound(str[i]);
          if (it == s.end()) {
            ans = *(s.begin()) + ans;
          } else {
            ans = *it + ans;
            smaller = true;
          }
        }
      }
      cout << ans << endl;
    }
  }
  return 0;
}
