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;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1;
  for (long long i = 0; i < t; i++) {
    long long n, k;
    cin >> n >> k;
    string s;
    cin >> s;
    vector<char> store;
    map<char, long long> count;
    for (long long x = 0; x < n; x++) {
      count[s[x]]++;
      if (count[s[x]] == 1) store.push_back(s[x]);
    }
    sort(store.begin(), store.end());
    string ans = "";
    if (k > n) {
      ans = s;
      while (ans.length() < k) {
        ans += store[0];
      }
    } else {
      for (long long x = k - 1; x >= 0; x--) {
        if (s[x] != store[store.size() - 1]) {
          for (long long y = 0; y < store.size(); y++) {
            if (store[y] > s[x]) {
              ans += store[y];
              break;
            }
          }
          break;
        } else
          ans += store[0];
      }
      reverse(ans.begin(), ans.end());
      string temp = ans;
      ans = s.substr(0, k - ans.length());
      ans += temp;
    }
    cout << ans << '\n';
  }
  return 0;
}
