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;
string s;
int seen[N];
string a;
int main() {
  cin >> n >> k >> s;
  for (int i = 0; i < ((int)(s).size()); i++) {
    int cur = s[i] - 'a';
    if (seen[cur]) continue;
    seen[cur] = 1;
    a += s[i];
  }
  sort((a).begin(), (a).end());
  if (k > n) {
    while (((int)(s).size()) < k) s += a[0];
    cout << s << '\n';
    return 0;
  }
  while (((int)(s).size()) > k) s.pop_back();
  n = ((int)(s).size());
  for (int i = k - 1; i >= 0; i--) {
    if (s[i] != a.back()) {
      int idx = lower_bound(begin(a), end(a), s[i]) - begin(a);
      s[i] = a[++idx];
      for (int j = i + 1; j < k; j++) s[j] = a[0];
      cout << s << '\n';
      return 0;
    }
  }
  assert(false);
  return 0;
}
