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;
int freq[200];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, k;
  cin >> n >> k;
  vector<char> vec;
  string str;
  cin >> str;
  for (int i = 0; i < n; i++) {
    if (freq[str[i]] == 0) {
      vec.push_back(str[i]);
      freq[str[i]]++;
    }
  }
  sort(vec.begin(), vec.end());
  if (k > n) {
    vector<char> b;
    for (int i = 0; i < n; i++) {
      b.push_back(str[i]);
    }
    for (int i = 1; i <= k - n; i++) {
      b.push_back(vec[0]);
    }
    int len = b.size();
    for (int i = 0; i < len; i++) {
      cout << b[i];
    }
    return 0;
  }
  int z = vec.size();
  vector<char> a;
  for (int i = 0; i < k; i++) {
    a.push_back(str[i]);
  }
  int flag = 0;
  for (int i = k - 1; i >= 0; i--) {
    if (a[i] != vec[z - 1]) {
      for (int j = 0; j < z - 1; j++) {
        if (vec[j] == a[i]) {
          a[i] = vec[j + 1];
          flag = 1;
          for (int m = i + 1; m < k; m++) {
            a[m] = vec[0];
          }
          break;
        }
      }
    }
    if (flag == 1) break;
  }
  for (int i = 0; i < k; i++) cout << a[i];
  return 0;
}
