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 n, d = -1, k;
string s1, s;
char _s[1000006];
char a[1000006];
string flus(string h) {
  int _n = h.length() - 1, du = 1;
  for (int i = _n; i >= 0 && du != 0; i--) {
    h[i] += 1;
    if (h[i] > d)
      h[i] = 0, du = 1;
    else
      du = 0;
  }
  return h;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> k;
  cin >> s;
  if (s.length() < k) {
    char minn = CHAR_MAX;
    cout << s;
    for (int i = 0; i < s.length(); i++) minn = min(minn, s[i]);
    for (int i = 1; i <= k - s.length(); i++) cout << (char)minn;
    return 0;
  }
  string s1 = s;
  for (int i = 0; i < n; i++) _s[i] = s[i];
  sort(_s, _s + n);
  a[++d] = _s[0];
  for (int i = 1; i < n; i++)
    if (_s[i] != _s[i - 1]) a[++d] = _s[i];
  for (int i = 0; i < s1.length(); i++)
    s1[i] = lower_bound(a, a + d, s1[i]) - a;
  string k1 = "";
  for (int i = 0; i < k; i++) k1 = k1 + (char)s1[i];
  s1 = k1;
  s1 = flus(s1);
  for (int i = 0; i < k; i++) cout << char(a[s1[i]]);
  return 0;
}
