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;
long long a[100007];
long long n, m, k, res, all, best, d, dd, A, B;
char c, minc, maxc, nextc;
long long have[1007];
string s;
char findn(char c) {
  c++;
  while (have[c] == 0) c++;
  return c;
}
int main() {
  cin >> n >> k >> s;
  minc = 'z' + 1;
  maxc = 'a' - 1;
  for (int i = 0; i < s.length(); i++) {
    c = s[i];
    have[c]++;
    if (c < minc) minc = c;
    if (c > maxc) maxc = c;
  }
  if (k > n) {
    cout << s;
    for (int i = 1; i <= k - n; i++) cout << minc;
    cout << endl;
    return 0;
  }
  if (k <= n) {
    for (int i = k - 1; i >= 0; i--) {
      if (s[i] != maxc) {
        nextc = findn(s[i]);
        for (int j = 0; j < i; j++) cout << s[j];
        cout << nextc;
        for (int j = 1; j <= k - i - 1; j++) cout << minc;
        cout << endl;
        return 0;
      }
    }
  }
  return 0;
}
