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 = INT_MAX;
const long long llinf = LONG_LONG_MAX;
int cnt[30];
int main() {
  int n, k;
  cin >> n >> k;
  string s;
  cin >> s;
  memset(cnt, 0, sizeof cnt);
  for (int i = 0; i < n; i++) {
    cnt[s[i] - 'a']++;
  }
  if (n < k) {
    cout << s;
    for (int i = n; i < k; i++) {
      for (int j = 0; j < 26; j++) {
        if (cnt[j]) {
          cout << (char)(j + 'a');
          break;
        }
      }
    }
    return 0;
  }
  int pos = 0;
  bool flag = 0;
  char ans[100005];
  for (int i = k - 1; i >= 0; i--) {
    for (int j = 0; j < 26; j++) {
      if (cnt[j] && j + 'a' > s[i]) {
        flag = 1;
        pos = i;
        ans[i] = j + 'a';
        break;
      }
    }
    if (flag) break;
  }
  for (int i = 0; i < pos; i++) cout << s[i];
  cout << (char)ans[pos];
  for (int i = pos + 1; i < k; i++) {
    for (int j = 0; j < 26; j++) {
      if (cnt[j]) {
        cout << (char)(j + 'a');
        break;
      }
    }
  }
  return 0;
}
