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 inf = 1e9;
const long long INF = 1e18;
int n, k;
string s;
set<char> st;
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  ;
  cin >> n >> k >> s;
  for (auto x : s) {
    st.insert(x);
  }
  char big = *st.rbegin();
  char small = *st.begin();
  if (k > n) {
    cout << s;
    for (int i = n; i < k; i++) {
      cout << small;
    }
  } else {
    int pos;
    for (int i = k - 1; i >= 0; i--) {
      if (s[i] != big) {
        pos = i;
        break;
      }
    }
    for (int i = 0; i < pos; i++) {
      cout << s[i];
    }
    cout << (*st.upper_bound(s[pos]));
    for (int i = pos + 1; i < k; i++) {
      cout << small;
    }
  }
  return 0;
}
