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;
const int N = 1e5 + 5;
int n, k;
char s[N];
char ch[N];
int g;
int main() {
  scanf("%d%d", &n, &k);
  scanf("%s", s + 1);
  char mx = s[1];
  for (int i = 1; i <= n; ++i) mx = max(mx, s[i]);
  if (k <= n) {
    for (g = 1; g < k; ++g) ch[g] = s[g];
    g = k - 1;
    if (s[k] == mx) {
      int j = k - 1;
      while (s[j] == mx) j--;
      char x = s[j];
      sort(s + 1, s + 1 + n);
      for (int i = 1; i <= n; ++i)
        if (s[i] > x) {
          x = s[i];
          break;
        }
      g = j;
      ch[g] = x;
      g++;
      for (; g <= k; ++g) ch[g] = s[1];
      printf("%s", ch + 1);
      return 0;
    }
    char x = s[k];
    sort(s + 1, s + 1 + n);
    for (int i = 1; i <= n; ++i)
      if (s[i] > x) {
        x = s[i];
        break;
      }
    ch[++g] = x;
    printf("%s", ch + 1);
    return 0;
  }
  if (k > n) {
    for (g = 1; g <= n; ++g) ch[g] = s[g];
    g = n;
    sort(s + 1, s + 1 + n);
    char x = s[1];
    for (g = n + 1; g <= k; ++g) ch[g] = x;
    printf("%s", ch + 1);
    return 0;
  }
  return 0;
}
