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;
string s, ans;
int n, t;
set<char> se;
void solve(int pl) {
  auto ed = se.end();
  --ed;
  if (s[pl] != *ed) {
    for (int i = 0; i < pl; i++) {
      ans.push_back(s[i]);
    }
    for (auto i : se) {
      if (i > s[pl]) {
        ans.push_back(i);
        break;
      }
    }
    char minch = *se.begin();
    for (int i = 0; i < t - pl - 1; i++) {
      ans.push_back(minch);
    }
  } else
    solve(pl - 1);
}
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> t >> s;
  for (int i = 0; i < s.size(); i++) {
    se.insert(s[i]);
  }
  auto ite = se.end();
  if (t > n) {
    for (int i = 0; i < n; i++) {
      ans.push_back(s[i]);
    }
    for (int i = 0; i < t - n; i++) ans.push_back(*se.begin());
  } else {
    solve(t - 1);
  }
  cout << ans << endl;
  return 0;
}
