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>
const long double eps = 0.00000001;
const long long MOD = 1e9 + 7;
using namespace std;
long long ans;
bool b;
string second, t;
int main() {
  fflush(stdin);
  cout << fixed, cout.precision(18);
  ios_base::sync_with_stdio(false);
  int i, j, n, m;
  cin >> n >> m >> second;
  set<char> se;
  for (i = 0; i < n; ++i) se.insert(second[i]);
  if (m <= n) {
    for (i = 0; i < m - 1; ++i) t += second[i];
    int cur = m - 1;
    while (1) {
      auto it = se.upper_bound(second[cur]);
      if (it == se.end())
        --cur, t.pop_back();
      else {
        t += (*it);
        break;
      }
    }
    while (t.size() < m) t += (*se.begin());
  } else {
    t = second;
    auto it = se.begin();
    for (i = n; i < m; ++i) t += (*it);
  }
  cout << t;
}
