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[235];
int main() {
  int n, k;
  cin >> n >> k;
  string s;
  cin >> s;
  set<char> q;
  int minn = 565, pos = -1;
  char c_min = 'z', c_max = 'a';
  for (int i = 0; i < n; i++) {
    a[s[i] - 'a'] = 1;
    c_min = min(c_min, s[i]);
    c_max = max(c_max, s[i]);
  }
  for (int i = 0; i < n; i++) {
    if (s[i] == c_max) {
      pos = i;
      break;
    }
  }
  if (k > n) {
    cout << s;
    int t = k - n;
    while (t--) cout << c_min;
    cout << endl;
  } else {
    int pos_1 = -1;
    char c2;
    for (int i = k - 1; i >= 0; i--) {
      if (pos_1 != -1) break;
      for (int j = 0; j <= 'z' - 'a'; j++) {
        if (a[j] == 1 && (j + 'a') > s[i]) {
          c2 = j + 'a';
          pos_1 = i;
          break;
        }
      }
    }
    for (int i = 0; i < pos_1; i++) {
      cout << s[i];
    }
    cout << c2;
    for (int i = pos_1 + 1; i < k; i++) {
      cout << c_min;
    }
    cout << endl;
  }
  return 0;
}
