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;
string s, hernea;
long long n, k, ver;
int main() {
  cin >> n >> k >> s;
  if (n >= k) {
    char lr = 'a';
    char tsio = 'z';
    for (long long i = 0; i < n; i++) {
      lr = lr < s[i] ? s[i] : lr;
      tsio = tsio > s[i] ? s[i] : tsio;
    }
    hernea = "";
    for (long long i = k - 1; i >= 0; i--) {
      if (ver == 1) {
        hernea = s[i] + hernea;
      } else {
        if (s[i] != lr) {
          char zz = lr;
          for (long long j = 0; j < n; j++) {
            if (s[j] < zz && s[j] > s[i]) {
              zz = s[j];
            }
          }
          ver = 1;
          hernea = zz + hernea;
        } else {
          hernea = tsio + hernea;
        }
      }
    }
  } else {
    char tsio = 'z';
    for (long long i = 0; i < n; i++) tsio = tsio > s[i] ? s[i] : tsio;
    hernea = s;
    for (long long i = n; i < k; i++) hernea += tsio;
  }
  cout << hernea;
}
