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>
const int MAXN = 1e5 + 10;
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  char s[MAXN];
  scanf("%s", s);
  bool vis[30];
  int pos = -1, poss;
  memset(vis, false, sizeof(vis));
  for (int i = 0; s[i]; ++i) vis[s[i] - 'a'] = true;
  for (int i = 0; s[i] && i <= k - 1; ++i) {
    for (int j = s[i] - 'a' + 1; j <= 25; ++j)
      if (vis[j]) {
        poss = j;
        pos = i;
        break;
      }
  }
  int f;
  for (int i = 0; i <= 25; ++i)
    if (vis[i]) {
      f = i;
      break;
    }
  if (pos == -1 && k <= n) {
    for (int i = 0; i <= 25; ++i)
      if (vis[i] && i > s[0] - 'a') {
        printf("%c", i + 'a');
        break;
      }
    for (int i = 0; i < k - 1; ++i) printf("%c", f + 'a');
    printf("\n");
  } else if (k <= n) {
    s[pos] = 'a' + poss;
    for (int i = 0; i <= pos; ++i) printf("%c", s[i]);
    for (int i = pos + 1; i < k; ++i) printf("%c", f + 'a');
    printf("\n");
  } else {
    printf("%s", s);
    for (int i = 0; i < k - n; ++i) printf("%c", f + 'a');
    printf("\n");
  }
}
