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;
vector<char> cnt;
map<char, char> tmp;
int main() {
  int n, k;
  cin >> n >> k;
  string str;
  cin >> str;
  for (auto it : str) cnt.push_back(it);
  sort(cnt.begin(), cnt.end());
  cnt.erase(unique(cnt.begin(), cnt.end()), cnt.end());
  for (auto it = cnt.begin(); it != cnt.end(); it++)
    tmp[*it] = cnt[(it - cnt.begin() + 1) % cnt.size()];
  string ans(str, 0, k);
  if (str.size() < k) {
    for (int i = str.size(); i < k; i++) ans += *cnt.begin();
    cout << ans << endl;
    return 0;
  }
  reverse(ans.begin(), ans.end());
  int flag = 1;
  for (auto& it : ans) {
    if (flag == 0)
      break;
    else {
      flag = 0;
      it = tmp[it];
      if (it == *cnt.begin()) flag = 1;
    }
  }
  for (auto it = ans.rbegin(); it != ans.rend(); it++) cout << *it;
  cout << endl;
  return 0;
}
