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;
long long int n, k, i, j, u = 0;
string s;
long long int a[26];
char c;
int main() {
  cin >> n >> k;
  cin >> s;
  for (i = 0; i < n; i++) a[s[i] - 'a'] = 1;
  if (k > n) {
    cout << s;
    for (i = 0; i < 26; i++) {
      if (a[i] != 0) break;
    }
    for (j = n; j < k; j++) cout << char(i + 'a');
    return 0;
  } else {
    for (i = k - 1; i >= 0; i--) {
      for (j = s[i] - 'a' + 1; j < 26; j++) {
        if (a[j] == 1) {
          u = 1;
          break;
        }
      }
      if (u == 1) break;
    }
    for (long long int t = 0; t < i; t++) cout << s[t];
    cout << char(j + 'a');
    for (j = 0; j < 26; j++) {
      if (a[j] == 1) {
        c = char(j + 'a');
        break;
      }
    }
    for (j = i + 1; j < k; j++) cout << c;
  }
}
