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, cd = 0;
  cin >> n >> k;
  string cad = "", aux, aux2;
  cin >> cad;
  aux = cad;
  aux2 = cad;
  set<int> st;
  set<int>::iterator it;
  sort(aux.begin(), aux.end());
  char Min = aux[0];
  for (int i = 0; i < cad.size(); ++i) {
    st.insert(cad[i]);
  }
  if (cad.size() < k) {
    for (int i = cad.size(); i <= k; ++i) {
      cad = cad + Min;
    }
  }
  if (k <= n) {
    int index = 0;
    for (int i = k - 1; i >= 0; --i) {
      it = upper_bound(st.begin(), st.end(), cad[i]);
      char x = *it;
      if (x > cad[i]) {
        cad[i] = x;
        index = i;
        break;
      }
    }
    for (int i = 0; i < k; ++i) {
      if (i <= index) {
        cout << cad[i];
      } else
        cout << Min;
    }
    cout << "\n";
  } else {
    for (int i = 0; i < k; ++i) {
      cout << cad[i];
    }
    cout << "\n";
  }
  return 0;
}
