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 find_next(vector<char> a, char c) {
  for (int i = 0; i < a.size(); i++) {
    if (a[i] == c) return a[i + 1];
  }
}
int main() {
  int n, k;
  cin >> n >> k;
  string s;
  cin >> s;
  set<char> st;
  set<char>::iterator it;
  vector<char> a;
  for (int i = 0; i < n; i++) st.insert(s[i]);
  for (auto c : st) a.push_back(c);
  string ans = "";
  char max_c = a[(int)a.size() - 1];
  char min_c = a[0];
  if (k > n) {
    ans += s;
    for (int i = n; i < k; i++) ans += min_c;
    cout << ans << endl;
    return 0;
  }
  int l = min(n, k);
  for (int i = 0; i < l; i++) ans += s[i];
  int i = l - 1;
  while (i >= 0) {
    if (s[i] == ans[i] && s[i] == max_c) {
      i--;
    } else if (s[i] == ans[i]) {
      char sub = find_next(a, s[i]);
      ans[i] = sub;
      for (int p = i + 1; p < k; p++) ans[p] = min_c;
      cout << ans << endl;
      return 0;
    }
  }
}
