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 n, k;
string val;
bool ex[26];
string ans = "";
bool isHigher(char c) {
  for (int i = c - 'a' + 1; i <= 26; i++)
    if (ex[i]) return true;
  return false;
}
void iterate() {
  int l = val.length();
  l--;
  while (isHigher(val[l]) == false) l--;
  ans = val.substr(0, l);
  for (int i = val[l] - 'a' + 1; i <= 26; i++)
    if (ex[i]) {
      ans.append(1, i + 'a');
      break;
    }
  for (int i = 0; i <= 26; i++)
    if (ex[i]) {
      int diff = k - ans.length();
      ans.append(diff, i + 'a');
      break;
    }
}
int main() {
  cin >> n >> k;
  for (int i = 0; i < 26; i++) ex[i] = false;
  cin >> val;
  for (int i = 0; i < n; i++) {
    ex[val[i] - 'a'] = true;
  }
  if (k > n) {
    ans += val;
    int e = 0;
    while (ex[e] == false) e++;
    ans.append(k - n, ('a' + e));
  } else if (k < n) {
    val = val.substr(0, k);
    iterate();
  } else {
    iterate();
  }
  cout << ans << endl;
  return 0;
}
