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;
int32_t main() {
  long long n, k;
  cin >> n >> k;
  string s;
  cin >> s;
  set<char> myset;
  for (auto x : s) myset.insert(x);
  vector<char> v;
  for (auto x : myset) v.push_back(x);
  if (k > n) {
    long long i = 0;
    string ans = "";
    while (i < n) {
      ans += s[i];
      i++;
    }
    while (i < k) {
      ans += v[0];
      i++;
    }
    cout << ans << "\n";
  } else {
    string ans = s.substr(0, k);
    long long i = k - 1;
    char value;
    while (i >= 0) {
      long long idx = upper_bound(v.begin(), v.end(), ans[i]) - v.begin();
      if (idx < v.size()) {
        value = v[idx];
        break;
      }
      i--;
    }
    ans[i] = value;
    i++;
    while (i < k) {
      ans[i] = v[0];
      i++;
    }
    cout << ans << "\n";
  }
}
