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;
char input[27];
int cnt = 0;
bool used[27];
char ans[100001];
int main() {
  string s;
  int n, k;
  int i, j;
  cin >> n >> k;
  cin >> s;
  for (i = 0; i < s.length(); i++)
    if (!used[s[i] - 'a' + 1]) used[s[i] - 'a' + 1] = true, input[cnt++] = s[i];
  sort(input, input + cnt);
  if (k > n) {
    for (i = 0; i < s.length(); i++) cout << s[i];
    for (i = s.length(); i < k; i++) cout << input[0];
    cout << endl;
  } else {
    int add = 1;
    i = k;
    while (i > 0) {
      int po;
      if (add) {
        for (j = 0; j < cnt; j++)
          if (input[j] == s[i - 1]) break;
        if (j == cnt - 1)
          ans[i] = input[0];
        else
          ans[i] = input[j + 1], add = 0;
      } else
        ans[i] = s[i - 1];
      i--;
    }
    for (i = 1; i <= k; i++) cout << ans[i];
    cout << endl;
  }
  return 0;
}
