#include <bits/stdc++.h>
using namespace std;
int n, m;
string s, t;
vector<char> vv;
int main() {
  cin >> n >> m;
  cin >> s;
  for (int i = 0; i < n; i++) vv.push_back(s[i]);
  sort(vv.begin(), vv.end());
  if (((int)(s).size()) < m) {
    t = s;
    for (int i = ((int)(s).size()); i < m; i++) t += vv[0];
    cout << t << endl;
    return 0;
  } else if (((int)(s).size()) == m) {
    t = s;
    int k = 1;
    while (1) {
      if (s[((int)(s).size()) - k] == vv[((int)(vv).size()) - 1]) {
        t[((int)(s).size()) - k] = vv[0];
        k++;
        continue;
      }
      int bb = upper_bound((vv).begin(), (vv).end(), t[((int)(s).size()) - k]) -
               vv.begin();
      t[((int)(s).size()) - k] = vv[bb];
      break;
    }
    cout << t << endl;
    return 0;
  } else {
    t = s.substr(0, m);
    int k = 1;
    while (1) {
      if (s[m - k] == vv[((int)(vv).size()) - 1]) {
        t[m - k] = vv[0];
        k++;
        continue;
      }
      int bb = upper_bound((vv).begin(), (vv).end(), t[m - k]) - vv.begin();
      t[m - k] = vv[bb];
      break;
    }
    cout << t << endl;
    return 0;
  }
}

### Prompt

Please create a solution in cpp 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>
using namespace std;
int n, m;
string s, t;
vector<char> vv;
int main() {
  cin >> n >> m;
  cin >> s;
  for (int i = 0; i < n; i++) vv.push_back(s[i]);
  sort(vv.begin(), vv.end());
  if (((int)(s).size()) < m) {
    t = s;
    for (int i = ((int)(s).size()); i < m; i++) t += vv[0];
    cout << t << endl;
    return 0;
  } else if (((int)(s).size()) == m) {
    t = s;
    int k = 1;
    while (1) {
      if (s[((int)(s).size()) - k] == vv[((int)(vv).size()) - 1]) {
        t[((int)(s).size()) - k] = vv[0];
        k++;
        continue;
      }
      int bb = upper_bound((vv).begin(), (vv).end(), t[((int)(s).size()) - k]) -
               vv.begin();
      t[((int)(s).size()) - k] = vv[bb];
      break;
    }
    cout << t << endl;
    return 0;
  } else {
    t = s.substr(0, m);
    int k = 1;
    while (1) {
      if (s[m - k] == vv[((int)(vv).size()) - 1]) {
        t[m - k] = vv[0];
        k++;
        continue;
      }
      int bb = upper_bound((vv).begin(), (vv).end(), t[m - k]) - vv.begin();
      t[m - k] = vv[bb];
      break;
    }
    cout << t << endl;
    return 0;
  }
}

```