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() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, m;
  while (cin >> n >> m) {
    string s;
    cin >> s;
    char minn = s[0], maxx = s[0];
    for (int i = 1; i < n; i++) {
      minn = min(minn, s[i]);
      maxx = max(maxx, s[i]);
    }
    if (n < m) {
      cout << s;
      for (int i = n; i < m; i++) cout << minn;
      cout << endl;
    } else {
      int i;
      for (i = m - 1; i >= 0; i--) {
        if (s[i] != maxx) break;
      }
      char lb = maxx;
      for (int j = 0; j < n; j++) {
        if (s[j] > s[i]) {
          lb = min(s[j], lb);
        }
      }
      s[i] = lb;
      for (int j = i + 1; j < m; j++) {
        s[j] = minn;
      }
      cout << s.substr(0, m) << endl;
    }
  }
}
