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;
char a[100005];
bool cnt[150], OK[100005];
int main() {
  int i, i0, n, m;
  while (scanf("%d %d", &n, &m) != EOF) {
    memset(OK, 0, sizeof(OK));
    memset(cnt, 0, sizeof(cnt));
    scanf("%s", a);
    int len = strlen(a);
    for (i = 0; i < n; i++) cnt[a[i]] = 1;
    char tmp;
    bool flag = 0;
    i0 = 125;
    for (i0; i0; i0--)
      if (cnt[i0]) break;
    if (n >= m) {
      for (i = m - 1; i; i--) {
        if (i0 > a[i] || OK[i]) OK[i - 1] = 1;
      }
    }
    if (n < m)
      for (i = 0; i < n; i++) OK[i] = 1;
    for (i = 0; i < m; i++) {
      for (i0 = 95; i0 < 125; i0++) {
        if (flag && cnt[i0]) {
          tmp = i0;
          break;
        }
        if (i < len && i0 == a[i] && cnt[i0] && OK[i]) {
          tmp = i0;
          break;
        } else if (i < len && i0 > a[i] && cnt[i0]) {
          tmp = i0;
          flag = 1;
          break;
        } else if (i >= len && cnt[i0]) {
          flag = 1;
          tmp = i0;
          break;
        }
      }
      printf("%c", tmp);
    }
    printf("\n");
  }
  return 0;
}
