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 s[101000], t[101000];
int n, k, mx, mn, fl[233];
int main() {
  cin >> n >> k >> s + 1;
  mx = 0, mn = 233;
  for (int i = 1; i <= n; ++i) {
    mx = max(mx, (int)s[i]);
    mn = min(mn, (int)s[i]);
    fl[s[i]] = 1;
  }
  if (n < k) {
    cout << s + 1;
    for (int i = n + 1; i <= k; ++i) printf("%c", mn);
    return 0;
  }
  for (int p = k;; --p)
    if (s[p] != mx) {
      ++s[p];
      for (; !fl[s[p]]; ++s[p])
        ;
      for (int i = 1; i <= p; ++i) printf("%c", s[i]);
      for (int i = p + 1; i <= k; ++i) printf("%c", mn);
      return 0;
    }
}
