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;
char s[100055];
bool have[255];
int main() {
  cin >> n >> k;
  scanf("%s", s);
  for (int i = 0; i < n; i++) {
    have[s[i]] = true;
  }
  if (k > n) {
    printf("%s", s);
    for (int i = 'a'; i <= 'z'; i++)
      if (have[i]) {
        for (int j = 0; j < k - n; j++) {
          printf("%c", i);
        }
        break;
      }
    puts("");
    return 0;
  }
  char lar = 'a';
  char sma = 'z';
  for (int i = 'a'; i <= 'z'; i++)
    if (have[i]) {
      lar = i;
    }
  for (int i = 'z'; i >= 'a'; i--)
    if (have[i]) {
      sma = i;
    }
  int pt = k - 1;
  while (s[pt] == lar) pt--;
  s[pt]++;
  while (!have[s[pt]]) s[pt]++;
  for (int i = pt + 1; i < n; i++) {
    s[i] = sma;
  }
  for (int i = 0; i < k; i++) printf("%c", s[i]);
  puts("");
  return 0;
}
