#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, k;
  string s;
  cin >> n >> k >> s;
  bool T[26];
  for (int i = 0; i < 26; ++i) T[i] = false;
  for (int i = 0; i < n; ++i) T[s[i] - 'a'] = true;
  int lcha = 48;
  for (int i = 0; i < 26; ++i) {
    if (T[i]) lcha = min(lcha, i);
  }
  if (k > n) {
    cout << s;
    for (int i = n; i < k; ++i) cout << (char)(lcha + 'a');
    cout << endl;
    return 0;
  }
  map<char, int> cti;
  map<int, char> itc;
  int idx = 0;
  for (int i = 0; i < 26; ++i) {
    if (T[i]) {
      cti[(char)(i + 'a')] = idx;
      itc[idx] = i + 'a';
      idx++;
    }
  }
  int A[k];
  for (int i = 0; i < k; ++i) {
    A[i] = (i < n ? cti[s[i]] : 0);
  }
  A[k - 1]++;
  for (int i = k - 1; i > 0; --i) {
    if (A[i] >= idx) {
      A[i - 1]++;
      A[i] -= idx;
    }
  }
  for (int i = 0; i < k; ++i) cout << itc[A[i]];
  cout << endl;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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>
using namespace std;
int main() {
  int n, k;
  string s;
  cin >> n >> k >> s;
  bool T[26];
  for (int i = 0; i < 26; ++i) T[i] = false;
  for (int i = 0; i < n; ++i) T[s[i] - 'a'] = true;
  int lcha = 48;
  for (int i = 0; i < 26; ++i) {
    if (T[i]) lcha = min(lcha, i);
  }
  if (k > n) {
    cout << s;
    for (int i = n; i < k; ++i) cout << (char)(lcha + 'a');
    cout << endl;
    return 0;
  }
  map<char, int> cti;
  map<int, char> itc;
  int idx = 0;
  for (int i = 0; i < 26; ++i) {
    if (T[i]) {
      cti[(char)(i + 'a')] = idx;
      itc[idx] = i + 'a';
      idx++;
    }
  }
  int A[k];
  for (int i = 0; i < k; ++i) {
    A[i] = (i < n ? cti[s[i]] : 0);
  }
  A[k - 1]++;
  for (int i = k - 1; i > 0; --i) {
    if (A[i] >= idx) {
      A[i - 1]++;
      A[i] -= idx;
    }
  }
  for (int i = 0; i < k; ++i) cout << itc[A[i]];
  cout << endl;
}

```