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;
int mi = 0, ma = 25;
bool exist[26];
char str[100005];
inline void init() {
  scanf("%d%d", &n, &k);
  memset(exist, false, sizeof(exist));
  scanf("%s", str + 1);
}
inline void solve() {
  for (int i = 1; i <= n; i++) exist[str[i] - 'a'] = true;
  for (mi = 0; !exist[mi]; mi++)
    ;
  for (ma = 25; !exist[ma]; ma--)
    ;
  if (n < k) {
    printf("%s", str + 1);
    for (int i = n + 1; i <= k; i++) putchar(mi + 'a');
    return;
  }
  int p;
  for (p = k; p && str[p] == ma + 'a'; p--)
    ;
  str[p]++;
  while (!exist[str[p] - 'a']) str[p]++;
  for (p = p + 1; p <= k; p++) str[p] = mi + 'a';
  str[k + 1] = 0;
  puts(str + 1);
}
int main() {
  init();
  solve();
  return 0;
}
