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;
char s[100005];
char ans[100005];
int main() {
  scanf("%d%d", &n, &k);
  scanf("%s", &s);
  vector<char> v;
  for (int i = 0; i < n; i++) {
    if (find(v.begin(), v.end(), s[i]) == v.end()) v.push_back(s[i]);
  }
  sort(v.begin(), v.end());
  if (k > n) {
    printf("%s", s);
    for (int i = 0; i < k - n; i++) printf("%c", v[0]);
  } else {
    int i;
    for (i = k - 1;; i--) {
      if (s[i] != v[v.size() - 1]) {
        int x = lower_bound(v.begin(), v.end(), s[i]) - v.begin();
        ans[i] = v[x + 1];
        break;
      } else
        ans[i] = v[0];
    }
    i--;
    for (; i >= 0; i--) ans[i] = s[i];
    printf("%s", ans);
  }
  return 0;
}
