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 N = 5e6 + 10;
int n, k, d = 200;
string s, Ans;
vector<int> I[30];
bool vis[30];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k >> s;
  Ans = s.substr(0, min(k, n));
  for (int i = 0; i < n; i++) I[s[i] - 'a'].push_back(i);
  for (int i = 0; i < n; i++) d = min(d, s[i] - 'a');
  if (n < k)
    for (int i = n + 1; i <= k; i++) Ans += (char)(d + 'a');
  else {
    for (int i = k - 1; i >= 0; i--) {
      int y = Ans[i] - 'a';
      if (vis[y]) continue;
      vis[y] = 1;
      for (int j = y + 1; j < 26; j++)
        if (I[j].size()) {
          Ans[i] = (char)(j + 'a');
          for (int u = 0; u < 26; u++) {
            for (int q = i + 1; q < k; q++) Ans[q] = (char)(d + 'a');
            return cout << Ans, 0;
          }
        }
    }
  }
  cout << Ans;
}
