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 n, k;
string str;
int freq[26];
int main() {
  scanf("%d %d", &n, &k);
  cin >> str;
  int i, j;
  for (i = 0; i < n; i++) freq[str[i] - 97]++;
  int xx = -1;
  for (i = 0; i < 26; i++) {
    if (freq[i] != 0) {
      xx = i;
      break;
    }
  }
  if (k > n) {
    cout << str;
    for (i = n + 1; i <= k; i++) {
      printf("%c", (char)(xx + 97));
    }
    return 0;
  } else {
    string str1 = "";
    for (i = 0; i < k; i++) str1 += str[i];
    for (i = k - 1; i >= 0; i--) {
      int xxx = str1[i] - 97;
      int co = -1;
      for (j = xxx + 1; j < 26; j++) {
        if (freq[j]) {
          co = j;
          break;
        }
      }
      if (co != -1) {
        str1[i] = (char)(co + 97);
        for (j = i + 1; j < k; j++) {
          str1[j] = (char)(xx + 97);
        }
        break;
      }
    }
    cout << str1 << "\n";
  }
}
