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;
int main() {
  int n, k;
  while (cin >> n >> k) {
    string ss;
    int vis[30] = {0};
    cin >> ss;
    for (int i = 0; i < n; i++) {
      vis[ss[i] - 'a'] = 1;
    }
    int flag = 0;
    int l, r;
    for (int i = 0; i < 26; i++) {
      if (vis[i]) {
        if (!flag) {
          l = i + 'a';
          flag = 1;
        }
        r = i + 'a';
      }
    }
    if (k > n) {
      cout << ss;
      for (int i = 0; i < k - n; i++) cout << char(l);
    } else {
      for (int i = k - 1; i >= 0; i--) {
        if (r > ss[i]) {
          cout << ss.substr(0, i);
          for (int j = ss[i] - 'a' + 1; j < 26; j++) {
            if (vis[j]) {
              cout << char(j + 'a');
              break;
            }
          }
          for (int j = i + 1; j < k; j++) {
            cout << char(l);
          }
          break;
        }
      }
    }
  }
  return 0;
}
