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 double PI = acos(-1);
const double eps = 1e-9;
int main() {
  long long i, j, k, l, p, q, r, x, y, z, n, m, c(0), t, h, mn = pow(2, 62) - 1,
                                                            mx = 0;
  cin >> n >> k;
  string s;
  cin >> s;
  map<char, bool> mp;
  long long sz = s.size() - 1;
  char cc, dd;
  vector<char> v;
  for (i = 0; i <= sz; i++) {
    mp[s[i]] = 1;
    v.push_back(s[i]);
  }
  sort(v.begin(), v.end());
  vector<char>::iterator ib, ie;
  dd = v[0];
  cc = v[v.size() - 1];
  x = -1;
  if (k > s.size()) {
    cout << s;
    for (i = 1; i <= k - s.size(); i++) cout << dd;
    return 0;
  }
  for (i = 0; i <= sz; i++) {
    if (i > k - 1) break;
    if (s[i] < cc) x = i;
  }
  bool f = 0;
  for (i = 0; i <= k - 1; i++) {
    if (f) {
      cout << dd;
      y = 0;
    } else if (x == i) {
      for (j = s[i] + 1; j <= 122; j++)
        if (mp[j] && j > s[i]) {
          char pp = j;
          cout << pp;
          break;
        }
      f = 1;
    } else
      cout << s[i];
  }
}
