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;
void init_ios() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
}
int n, k;
bool vis[30];
string s;
int main() {
  init_ios();
  cin >> n >> k >> s;
  int l = s.size();
  for (int i = 0; i < l; ++i) vis[s[i] - 'a'] = true;
  char wtf1, wtf2;
  for (int i = 0; i < 30; ++i)
    if (vis[i]) {
      wtf1 = (char)(i + 'a');
      break;
    }
  for (int i = 30; i >= 0; --i)
    if (vis[i]) {
      wtf2 = (char)(i + 'a');
      break;
    }
  if (k > l) {
    cout << s;
    for (int i = 0; i < k - l; ++i) cout << wtf1;
    cout << "\n";
    return 0;
  }
  string res = s.substr(0, k);
  for (int i = k - 1; i >= 0; --i)
    if (res[i] != wtf2) {
      char co;
      for (int j = (int)(res[i] - 'a') + 1; j < 30; ++j)
        if (vis[j]) {
          co = (char)(j + 'a');
          break;
        }
      res[i] = co;
      for (int j = i + 1; j < k; ++j) res[j] = wtf1;
      break;
    }
  cout << res << "\n";
}
