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 b, c, i;
  string a;
  map<char, char> d;
  map<char, char>::iterator it;
  scanf("%d %d", &b, &c);
  cin >> a;
  for (i = 0; i < b; i++) {
    if (d.find(a[i]) == d.end()) d[a[i]] = a[i];
  }
  if (b >= c) {
    b = c;
    for (b--; b >= 0; b--) {
      it = d.find(a[b]);
      for (it++; it != d.end(); it++) {
        a[b] = it->first;
        break;
      }
      if (it != d.end()) {
        break;
      }
    }
  } else
    b--;
  b++;
  for (i = 0; i < b && (d.size() != 1 || b < c); i++) {
    printf("%c", a[i]);
  }
  for (; b < c; b++) {
    printf("%c", d.begin()->first);
  }
}
