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 main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long n, m = 0, i, j = 1, l = 0, r = 0, x = 0, u = 0, y = 1, k = 0,
               ans = 4000000000000000007;
  cin >> n >> k;
  string s;
  cin >> s;
  set<char> p;
  for (i = (0); i <= (n - 1); i++) l = max(l, (long long)s[i]), p.insert(s[i]);
  if (k <= n) {
    char a[k];
    for (i = (k - 1); i >= (1); i--) {
      if (s[i] < l) break;
    }
    for (auto e : p) {
      if (e > s[i]) {
        a[i] = e;
        break;
      }
    }
    for (j = (0); j <= (i - 1); j++) {
      a[j] = s[j];
    }
    for (j = (i + 1); j <= (k - 1); j++) a[j] = *p.begin();
    for (i = (0); i <= (k - 1); i++) cout << a[i];
    return 0;
  } else {
    char a[k];
    for (i = (0); i <= (n - 1); i++) a[i] = s[i];
    for (j = (n); j <= (k - 1); j++) a[j] = *(p.begin());
    for (i = (0); i <= (k - 1); i++) cout << a[i];
    return 0;
  }
}
