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 inf = 1e9 + 7;
const int N = 1e6 + 5;
string s, t = "";
int a[30];
vector<char> v;
int main() {
  int n, k;
  cin >> n >> k;
  cin >> s;
  for (int i = 0; i < s.size(); i++) {
    a[s[i] - 'a'] = 1;
  }
  for (int i = 0; i < 26; i++)
    if (a[i]) v.push_back(char(i + 'a'));
  for (int i = 0; i < min(k, n); i++) t += s[i];
  if (n < k) {
    for (int i = n; i < k; i++) t += v[0];
  } else {
    for (int i = k - 1; i >= 0; i--) {
      if (t[i] == v.back()) continue;
      for (int j = 0; j < v.size(); j++)
        if (v[j] > t[i]) {
          t[i] = v[j];
          break;
        }
      for (int j = i + 1; j < k; j++) t[j] = v[0];
      break;
    }
  }
  cout << t << endl;
  return 0;
}
