#include <bits/stdc++.h>
using namespace std;
const int N = 200005;
const int inf = 1000 * 1000 * 1000;
const int mod = 1000 * 1000 * 1000 + 7;
int n, k;
string s;
bool used[N];
int main() {
  cin >> n >> k;
  cin >> s;
  int mx = -1;
  for (int i = 0; i < n; i++) {
    int x = s[i] - 'a';
    used[x] = true;
  }
  if (k > n) {
    for (int i = 0; i < n; i++) cout << s[i];
    int x;
    for (int i = 0; i < 26; i++) {
      if (used[i]) {
        x = i;
        break;
      }
    }
    for (int i = n; i < k; i++) cout << char(x + 'a');
    cout << endl;
    return 0;
  }
  int x;
  for (int i = 25; i >= 0; i--) {
    if (used[i]) {
      x = i;
      break;
    }
  }
  int id = -1;
  for (int i = k - 1; i >= 0; i--) {
    if (s[i] - 'a' != x) {
      id = i;
      break;
    }
  }
  for (int i = 0; i < id; i++) cout << s[i];
  for (int i = 0; i < 26; i++) {
    if (used[i] == false) continue;
    if (i > s[id] - 'a') {
      cout << char(i + 'a');
      break;
    }
  }
  for (int i = 0; i < 26; i++) {
    if (used[i]) {
      x = i;
      break;
    }
  }
  for (int i = id + 1; i < k; i++) cout << char(x + 'a');
  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>
using namespace std;
const int N = 200005;
const int inf = 1000 * 1000 * 1000;
const int mod = 1000 * 1000 * 1000 + 7;
int n, k;
string s;
bool used[N];
int main() {
  cin >> n >> k;
  cin >> s;
  int mx = -1;
  for (int i = 0; i < n; i++) {
    int x = s[i] - 'a';
    used[x] = true;
  }
  if (k > n) {
    for (int i = 0; i < n; i++) cout << s[i];
    int x;
    for (int i = 0; i < 26; i++) {
      if (used[i]) {
        x = i;
        break;
      }
    }
    for (int i = n; i < k; i++) cout << char(x + 'a');
    cout << endl;
    return 0;
  }
  int x;
  for (int i = 25; i >= 0; i--) {
    if (used[i]) {
      x = i;
      break;
    }
  }
  int id = -1;
  for (int i = k - 1; i >= 0; i--) {
    if (s[i] - 'a' != x) {
      id = i;
      break;
    }
  }
  for (int i = 0; i < id; i++) cout << s[i];
  for (int i = 0; i < 26; i++) {
    if (used[i] == false) continue;
    if (i > s[id] - 'a') {
      cout << char(i + 'a');
      break;
    }
  }
  for (int i = 0; i < 26; i++) {
    if (used[i]) {
      x = i;
      break;
    }
  }
  for (int i = id + 1; i < k; i++) cout << char(x + 'a');
  cout << endl;
  return 0;
}

```