#include <bits/stdc++.h>
#pragma warning(disable : 4996)
using namespace std;
int main() {
  long long n, k;
  cin >> n >> k;
  vector<char> s(n);
  set<char> a;
  for (long long i = 0; i < n; i++) {
    cin >> s[i];
    a.insert(s[i]);
  }
  vector<char> snew(a.begin(), a.end());
  sort(snew.begin(), snew.end());
  long long l = snew.size();
  vector<char> t(k, snew[0]);
  for (long long i = 0; i < min(k, n); i++) {
    t[i] = s[i];
  }
  if (k > n) {
    for (long long i = 0; i < k; i++) cout << t[i];
    return 0;
  }
  t[k - 1] =
      snew[(find(snew.begin(), snew.end(), t[k - 1]) - snew.begin() + 1) % l];
  for (long long i = k - 1; t[i] < s[i] && i > 0; i--) {
    t[i - 1] =
        snew[(find(snew.begin(), snew.end(), t[i - 1]) - snew.begin() + 1) % l];
  }
  for (long long i = 0; i < k; i++) cout << t[i];
  cout << endl;
  return 0;
}

### Prompt

Your task is to create a CPP solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
#pragma warning(disable : 4996)
using namespace std;
int main() {
  long long n, k;
  cin >> n >> k;
  vector<char> s(n);
  set<char> a;
  for (long long i = 0; i < n; i++) {
    cin >> s[i];
    a.insert(s[i]);
  }
  vector<char> snew(a.begin(), a.end());
  sort(snew.begin(), snew.end());
  long long l = snew.size();
  vector<char> t(k, snew[0]);
  for (long long i = 0; i < min(k, n); i++) {
    t[i] = s[i];
  }
  if (k > n) {
    for (long long i = 0; i < k; i++) cout << t[i];
    return 0;
  }
  t[k - 1] =
      snew[(find(snew.begin(), snew.end(), t[k - 1]) - snew.begin() + 1) % l];
  for (long long i = k - 1; t[i] < s[i] && i > 0; i--) {
    t[i - 1] =
        snew[(find(snew.begin(), snew.end(), t[i - 1]) - snew.begin() + 1) % l];
  }
  for (long long i = 0; i < k; i++) cout << t[i];
  cout << endl;
  return 0;
}

```