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 long long N = 35000;
int main() {
  long long x, i, j, k, z, w, n, m, d, p, h, r, l, y, t;
  string s, v;
  cin >> n >> m >> s;
  set<char> st;
  for (auto x : s) st.insert(x);
  v = s;
  sort(v.begin(), v.end());
  if (n < m)
    cout << s + string(m - n, v[0]);
  else {
    string ans = s.substr(0, m);
    for (i = m - 1; i >= 0; i--) {
      auto it = st.upper_bound(ans[i]);
      if (it == st.end())
        continue;
      else {
        ans[i] = *it;
        for (j = i + 1; j < m; j++) ans[j] = *st.begin();
        ans[j] = *st.begin();
        cout << ans;
        break;
      }
    }
  }
}
