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 maxn = 1e5 + 10;
int n, k;
string s, ans;
bool mark[26];
vector<char> v;
int32_t main() {
  cin >> n >> k >> s;
  for (char c : s) {
    if (!mark[c - 'a']) {
      mark[c - 'a'] = 1;
      v.push_back(c);
    }
  }
  sort(v.begin(), v.end());
  if (n < k) {
    cout << s;
    for (int i = int(n); i < int(k); i++) {
      cout << v[0];
    }
    cout << endl;
    return 0;
  }
  for (int i = int(0); i < int(k); i++) {
    ans.push_back(s[i]);
  }
  int j = k - 1;
  while (j >= 0 && ans[j] == v.back()) {
    ans[j] = v[0];
    j--;
  }
  if (j < 0) return -1;
  for (int i = int(ans[j] - 'a' + 1); i < int(26); i++) {
    if (mark[i]) {
      ans[j] = char(i + 'a');
      break;
    }
  }
  cout << ans << endl;
}
