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 a[100001];
int main() {
  int n, k, ma = 0, mi = INT_MAX, kq = INT_MAX, dd;
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) {
    char b;
    cin >> b;
    a[i] = b - 'a';
    mi = min(mi, a[i]);
  }
  for (int i = 1; i <= n; ++i) {
    ma = max(ma, a[i]);
  }
  if (n >= k) {
    for (int i = k; i >= 1; --i) {
      int tt = 0;
      if (a[i] != ma)
        for (int j = 1; j <= n; ++j)
          if (a[j] > a[i]) {
            kq = min(kq, a[j]);
            dd = i;
            tt = 1;
          }
      if (tt == 1) break;
    }
    for (int i = 1; i <= k; ++i) {
      string x;
      if (i < dd) x = (a[i] + 'a');
      if (i == dd) x = (kq + 'a');
      if (i > dd) x = (mi + 'a');
      cout << x;
    }
  } else
    for (int i = 1; i <= n; ++i) {
      string x;
      x = (a[i] + 'a');
      cout << x;
    }
  for (int i = 1; i <= k - n; ++i) {
    string x;
    x = (mi + 'a');
    cout << x;
  }
}
