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[100005], ss[30];
map<char, int> mp, mp1;
int main() {
  int n, t, m = 0;
  scanf("%d%d", &n, &t);
  scanf("%s", s);
  for (int i = 0; i < n; i++) {
    if (mp[s[i]] == 0) ss[m++] = s[i];
    mp[s[i]] = 1;
  }
  sort(ss, ss + m);
  for (int i = 0; i < m; i++) mp1[ss[i]] = i + 1;
  if (n < t) {
    for (int i = n; i < t; i++) s[i] = ss[0];
  } else {
    for (int i = t; i <= n - 1; i++) s[i] = '\0';
    for (int i = t - 1; i >= 0; i--) {
      if (mp1[s[i]] != m) {
        s[i] = ss[mp1[s[i]]];
        for (int j = i + 1; j < t; j++) s[j] = ss[0];
        break;
      }
    }
  }
  printf("%s", s);
  return 0;
}
