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() {
  int n, k;
  string S;
  cin >> n >> k >> S;
  set<char> exist;
  for (int(i) = 0; (i) < (int)(S.size()); (i)++) exist.insert(S[i]);
  vector<char> order;
  for (auto x : exist) order.push_back(x);
  if (n < k) {
    for (int(i) = 0; (i) < (int)(k - n); (i)++) S.push_back(order[0]);
    cout << S << endl;
  } else {
    S = S.substr(0, k);
    for (int(i) = (int)(S.size()) - 1; i >= 0; i--) {
      if (S[i] < order.back()) {
        for (int(k) = 0; (k) < (int)(order.size()); (k)++) {
          if (order[k] > S[i]) {
            S[i] = order[k];
            cout << S << endl;
            return 0;
          }
        }
      } else {
        S[i] = order[0];
      }
    }
  }
  return 0;
}
