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, kk;
  cin >> n;
  cin >> kk;
  bool avail[26];
  for (int i = 0; i < 26; i++) avail[i] = false;
  string s;
  cin >> s;
  for (int i = 0; i < s.size(); i++) avail[s[i] - 'a'] = true;
  string t;
  t = s.substr(0, kk);
  if (kk > t.size()) {
    int sm = -1;
    for (int i = 0; i < 26; i++)
      if (avail[i]) {
        sm = i;
        break;
      }
    while (kk > t.size()) {
      t += (sm + 'a');
    }
    cout << t << endl;
    return 0;
  }
  for (int i = t.size() - 1; i >= 0; i--) {
    bool found = false;
    int ind = -1;
    for (int j = t[i] - 'a' + 1; j < 26; j++) {
      if (avail[j] == true && s[i] != 'a' + j) {
        found = true;
        ind = j;
        break;
      }
    }
    if (found) {
      t[i] = 'a' + ind;
      int smallest = -1;
      for (int k = 0; k < 26; k++) {
        if (avail[k] == true) {
          smallest = k;
          break;
        }
      }
      for (int l = i + 1; l < t.size(); l++) {
        t[l] = smallest + 'a';
      }
      break;
    }
  }
  cout << t << endl;
  return 0;
}
