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 nxt[26];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  string x;
  cin >> n >> k >> x;
  for (auto c : x) nxt[c - 'a'] = 1;
  if (k <= n) {
    string res = "";
    int last = -1;
    for (int i = k - 1; i >= 0; i--) {
      if (last == -1)
        for (auto j = x[i] - 'a' + 1; j < (long long)(26); j++) {
          if (nxt[j]) {
            last = i;
            break;
          }
        }
    }
    for (auto i = 0; i < (long long)(last); i++) res.push_back(x[i]);
    int first = -1;
    for (auto i = 0; i < (long long)(26); i++) {
      if (nxt[i] && first == -1) first = i;
      if (i + 'a' > x[last] && nxt[i]) {
        res.push_back(i + 'a');
        break;
      }
    }
    for (auto i = ((int)((res).size())); i < (long long)(k); i++)
      res.push_back(first + 'a');
    cout << res << '\n';
  } else {
    cout << x;
    int first = -1;
    for (auto i = 0; i < (long long)(26); i++) {
      if (first == -1 && nxt[i]) first = i;
    }
    for (auto i = n; i < (long long)(k); i++) cout << (char)(first + 'a');
  }
  return 0;
}
