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 main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  set<char> s;
  string a1;
  cin >> a1;
  for (char i : a1) s.insert(i);
  int len = s.size();
  vector<char> v;
  set<char>::iterator it;
  for (it = s.begin(); it != s.end(); ++it) {
    v.push_back(*it);
  }
  map<char, char> make_pair;
  for (int i = 0; i < len; i++) {
    make_pair[v[i]] = v[(i + 1) % len];
  }
  int toNote = v[len - 1];
  int startpos = k - 1;
  if (k <= n) {
    string a2(a1, 0, k);
    for (int i = 0; i <= startpos; i++) {
      a2[i] = a1[i];
    }
    for (int i = startpos; i >= 0; i--) {
      if (a1 >= a2 && a2[i] != toNote) {
        a2[i] = make_pair[a2[i]];
        break;
      } else if (a1 >= a2 && a2[i] == toNote) {
        a2[i] = make_pair[a2[i]];
        a2[i - 1] = make_pair[a2[i - 1]];
        i--;
      } else {
        break;
      }
    }
    cout << a2 << '\n';
  } else {
    string a2(a1, 0, n);
    cout << a2;
    for (int i = 0; i < k - n; i++) {
      cout << v[0];
    }
    cout << '\n';
  }
  return 0;
}
