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;
const int maxn = 1e5 + 100;
int ax[maxn], vis[maxn], pos[maxn];
int n, k, A, B;
char s[maxn], s1[maxn];
vector<char> ac;
int main() {
  cin >> n >> k;
  cin >> s;
  int cnt = 0;
  int len = strlen(s);
  for (int i = 0; i < len; i++) {
    if (!vis[s[i] - 'a']) ac.push_back(s[i] - 'a'), vis[s[i] - 'a'] = 1;
  }
  sort(ac.begin(), ac.end());
  for (int i = 0; i < ac.size(); i++) pos[ac[i]] = i + 1;
  int m = ac.size();
  if (n >= k) {
    int ma = k;
    while (pos[s[k - 1] - 'a'] == m) {
      s1[k - 1] = (char)(ac[0] + 'a');
      k--;
    }
    s1[k - 1] = (char)(ac[pos[s[k - 1] - 'a']] + 'a');
    for (int i = 0; i < k - 1; i++) cout << s[i];
    for (int i = k - 1; i < ma; i++) cout << s1[i];
  } else {
    cout << s;
    for (int i = 0; i < k - n; i++) cout << (char)(ac[0] + 'a');
  }
  return 0;
}
