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;
  int tmp = k;
  string x;
  cin >> x;
  int mark[26];
  for (int j = 0; j < 26; j++) mark[j] = 0;
  for (int j = 0; j < x.length(); j++) {
    mark[x[j] - 'a'] = 1;
  }
  string ans = "";
  if (k > n) {
    ans = x;
    int loc = 0;
    for (int j = 0; j < 26; j++) {
      if (mark[j] == 1) {
        loc = j;
        break;
      }
    }
    int rem = k - n;
    while (rem--) {
      ans += ('a' + loc);
    }
    cout << ans << endl;
    return 0;
  }
  int ploc = 0;
  for (int j = 25; j >= 0; j--)
    if (mark[j] == 1) {
      ploc = j;
      break;
    }
  k--;
  while (k >= 0) {
    if (!(x[k] == (char)('a' + ploc))) break;
    k--;
  }
  for (int j = (int)(x[k] - 'a') + 1; j < 26; j++)
    if (mark[j] == 1) {
      ploc = j;
      break;
    }
  int myloc = k + 1;
  int xploc;
  for (int j = 0; j < 26; j++)
    if (mark[j] == 1) {
      xploc = j;
      break;
    }
  while (myloc < tmp) {
    ans += (char)('a' + xploc);
    myloc++;
  }
  k--;
  ans = (char)(ploc + 'a') + ans;
  while (k >= 0) {
    ans = x[k] + ans;
    k--;
  }
  cout << ans << "\n";
  return 0;
}
