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>
#pragma GCC optimize("O3")
using namespace std;
stringstream output;
inline void eOP();
inline void solve() {
  int n, k;
  cin >> n >> k;
  string s;
  cin >> s;
  bool* letters = new bool[26]();
  for (int i = 0; i < n; ++i) {
    letters[s[i] - 'a'] = true;
  }
  vector<char> decode;
  for (int i = 0; i < 26; ++i) {
    if (letters[i]) {
      decode.push_back(i + 'a');
    }
  }
  unordered_map<char, int> encode;
  for (int i = 0; i < decode.size(); ++i) {
    encode[decode[i]] = i;
  }
  vector<int> t;
  t.reserve(k);
  const size_t L = min(n, k);
  for (int i = 0; i < L; ++i) {
    t.push_back(encode[s[i]]);
  }
  t.resize(k);
  if (k <= n) {
    int i = k - 1;
    ++t[i];
    while (t[i] == decode.size()) {
      t[i] = 0;
      --i;
      ++t[i];
    }
  }
  for (int code : t) {
    output << decode[code];
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  size_t t = 1;
  for (int i = 0; i < t; ++i) {
    solve();
  }
  eOP();
}
inline void eOP() {
  cout << output.str();
  output.str(string());
  exit(0);
}
