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;
char c[100005];
set<char> s;
char d[100005];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  long long int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> c[i];
    s.insert(c[i]);
  }
  char u = *s.begin();
  char v = *(++s.begin());
  char t = *(--s.end());
  long long int q = 0;
  for (auto i : s) {
    q++;
    d[q] = i;
  }
  if (k > n) {
    for (int i = 1; i <= n; i++) cout << c[i];
    for (int i = n + 1; i <= k; i++) cout << d[1];
    return 0;
  }
  long long int j = 1;
  for (int i = k; i >= 1; i--) {
    if (c[i] != t) {
      j = i;
      break;
    }
  }
  for (int i = 1; i <= j - 1; i++) {
    cout << c[i];
  }
  for (int i = 1; i <= q; i++) {
    if (d[i] == c[j]) {
      cout << d[i + 1];
      break;
    }
  }
  for (int i = j + 1; i <= k; i++) {
    cout << d[1];
  }
  return 0;
}
