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;
map<char, bool> vst;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  string s, t = "";
  for (int i = 0; i < k; ++i) t.insert(t.begin() + i, '0');
  cin >> s;
  vector<char> v;
  for (int i = 0; i < s.size(); ++i) {
    if (!vst[s[i]]) {
      v.push_back(s[i]);
      vst[s[i]] = 1;
    }
  }
  sort(v.begin(), v.end());
  char mx = v[v.size() - 1], mn = v[0];
  if (n >= k) {
    int j = k - 1;
    while (s[j] == mx) {
      --j;
    }
    int x = upper_bound(v.begin(), v.end(), s[j]) - v.begin();
    t[j] = v[x];
    for (int i = 0; i < j; ++i) t[i] = s[i];
    for (int i = j + 1; i < k; ++i) t[i] = mn;
  } else {
    for (int i = 0; i < n; ++i) t[i] = s[i];
    for (int i = n; i < k; ++i) t[i] = mn;
  }
  cout << t << "\n";
  return 0;
}
