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 __precision = 13;
using ll = long long;
ll solve(ll n, ll k, ll A, ll B) {
  if (n < k) return (n - 1) * A;
  ll rem = n % k;
  return min((n - 1) * A, rem * A + B + solve(n / k, k, A, B));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.precision(__precision);
  cout.setf(ios_base::fixed);
  int n, k;
  cin >> n >> k;
  string s;
  cin >> s;
  set<char> st(s.begin(), s.end());
  string ans = s.substr(0, k);
  if (k <= n) {
    bool change = true;
    int cur = k - 1;
    while (change) {
      auto c = st.upper_bound(ans[cur]);
      if (c != st.end()) {
        ans[cur] = *c;
        for (int i = cur + 1; i < k; ++i) ans[i] = *st.begin();
        change = false;
      }
      --cur;
    }
  } else {
    while (ans.size() < k) {
      ans.push_back(*st.begin());
    }
  }
  cout << ans;
  return 0;
}
