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 maxn = 1e5 + 7;
char s[40];
int id = 0;
bool mark[40];
char op[maxn];
char ans[maxn];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  scanf("%s", op);
  int len = strlen(op) - 1;
  for (int i = 0; i <= len; ++i) {
    if (mark[op[i] - 'a']) continue;
    mark[op[i] - 'a'] = 1;
    s[id++] = op[i];
  }
  sort(s, s + id);
  if (k > len + 1) {
    printf("%s", op);
    for (int i = 1; i <= k - len - 1; ++i) {
      printf("%c", s[0]);
    }
    puts("");
  } else if (k < len + 1) {
    int m = k - 1;
    while (op[m] == s[id - 1]) m--;
    for (int i = 0; i <= k - 1; ++i) {
      if (i < m)
        printf("%c", op[i]);
      else if (i == m) {
        int inx = lower_bound(s, s + id, op[i]) - s;
        printf("%c", s[inx + 1]);
      } else
        printf("%c", s[0]);
    }
    puts("");
  } else {
    int m = len;
    while (op[m] == s[id - 1]) m--;
    for (int i = 0; i <= len; ++i) {
      if (i < m)
        printf("%c", op[i]);
      else if (i == m) {
        int inx = lower_bound(s, s + id, op[i]) - s;
        printf("%c", s[inx + 1]);
      } else
        printf("%c", s[0]);
    }
    puts("");
  }
  return 0;
}
