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;
  cin >> n >> k;
  string s;
  cin >> s;
  int tmp[256] = {0};
  set<char> S;
  for (auto it : s) tmp[it] = 1;
  for (int i = 0; i < 256; ++i)
    if (tmp[i] == 1) S.insert(i);
  if (k > n)
    cout << s << string(k - n, *S.begin()) << endl;
  else {
    string ans(s, 0, k);
    char largest = *S.rbegin();
    char smallest = *S.begin();
    for (int i = k - 1; i >= 0; i--) {
      if (ans[i] == largest)
        ans[i] = smallest;
      else {
        auto it = S.find(ans[i]);
        ans[i] = *(++it);
        break;
      }
    }
    cout << ans << endl;
  }
  return 0;
}
