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 a[30];
int n, k;
string str, s1;
char xl[30], mm;
int main() {
  cin >> n >> k;
  cin >> str;
  for (int i = 0; i < n; i++) {
    int tmp = str[i] - 'a' + 1;
    a[tmp] = 1;
  }
  int cnt = 0;
  for (int i = 0; i <= 29; i++) {
    if (a[i] == 1) {
      char tmp = i - 1 + 'a';
      xl[++cnt] = tmp;
      mm = tmp;
    }
  }
  if (k > n) {
    cout << str;
    for (int i = 1; i <= k - n; i++) cout << xl[1];
    cout << endl;
  } else {
    s1 = str;
    for (int i = k - 1; i >= 0; i--) {
      if (s1[i] == mm)
        s1[i] = xl[1];
      else {
        int tmp = s1[i] - 'a' + 1;
        for (int j = tmp + 1;; j++) {
          if (a[j] == 1) {
            s1[i] = j + 'a' - 1;
            break;
          }
        }
        break;
      }
    }
    for (int i = 0; i < k; i++) cout << s1[i];
    cout << endl;
  }
  return 0;
}
