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 inf = (1e9) + 7;
int mod = 998244353;
int chars[26];
int front[100005];
int main() {
  int n, t;
  scanf("%d %d", &n, &t);
  string s;
  cin >> s;
  char mx = 'a';
  char mn = 'z';
  for (int i = 0; i < s.size(); ++i) {
    chars[s[i] - 'a'] = 1;
    mx = max(mx, s[i]);
    mn = min(mn, s[i]);
  }
  if (t > n) {
    string ans = s;
    for (int i = 0; i < (t - n); ++i) ans += mn;
    cout << ans << endl;
    return 0;
  }
  bool b = true;
  for (int i = t - 1; i >= 0; --i) {
    b &= (s[i] == mx);
    front[i] = (b);
  }
  string ans = "";
  bool ja = false;
  for (int i = 0; i < t; ++i) {
    if (ja) {
      ans += mn;
    } else {
      if (i != t - 1 and front[i + 1]) {
        char c = s[i];
        for (int i = c - 'a' + 1; i < 26; ++i) {
          if (chars[i]) {
            c = 'a' + i;
            break;
          }
        }
        ans += c;
        ja = true;
      } else if (i != t - 1) {
        ans += s[i];
      } else {
        char c = s[i];
        for (int i = c - 'a' + 1; i < 26; ++i) {
          if (chars[i]) {
            c = 'a' + i;
            break;
          }
        }
        ans += c;
        ja = true;
      }
    }
  }
  cout << ans << endl;
}
