#include <bits/stdc++.h>
static char v[100000 + 2];
static char have[26];
int main(void) {
  int n, k, i, j, is_bigger, found, fill;
  scanf("%d %d\n", &n, &k);
  fgets(v, sizeof v, stdin);
  for (i = 0; i < n; i++) {
    if ('a' <= v[i] && v[i] <= 'z') {
      have[v[i] - 'a'] = 1;
    }
  }
  if (k > n) {
    fill = -1;
    for (i = 0; fill == -1 && i < 26; i++) {
      if (have[i]) {
        fill = 'a' + i;
      }
    }
    for (i = 0; i < n; i++) {
      putchar(v[i]);
    }
    for (; i < k; i++) {
      putchar(fill);
    }
  } else {
    is_bigger = 0;
    for (i = k - 1; !is_bigger && i >= 0; i--) {
      for (j = v[i] - 'a' + 1; !is_bigger && j < 26; j++) {
        if (have[j]) {
          is_bigger = 1;
          fill = 'a' + j;
        }
      }
    }
    for (j = 0; j < i + 1; j++) {
      putchar(v[j]);
    }
    putchar(fill);
    for (i = 0; i < 26; i++) {
      if (have[i]) {
        fill = 'a' + i;
        break;
      }
    }
    for (j++; j < k; j++) {
      putchar(fill);
    }
  }
  return 0;
}

### Prompt

In cpp, your task is to solve 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>
static char v[100000 + 2];
static char have[26];
int main(void) {
  int n, k, i, j, is_bigger, found, fill;
  scanf("%d %d\n", &n, &k);
  fgets(v, sizeof v, stdin);
  for (i = 0; i < n; i++) {
    if ('a' <= v[i] && v[i] <= 'z') {
      have[v[i] - 'a'] = 1;
    }
  }
  if (k > n) {
    fill = -1;
    for (i = 0; fill == -1 && i < 26; i++) {
      if (have[i]) {
        fill = 'a' + i;
      }
    }
    for (i = 0; i < n; i++) {
      putchar(v[i]);
    }
    for (; i < k; i++) {
      putchar(fill);
    }
  } else {
    is_bigger = 0;
    for (i = k - 1; !is_bigger && i >= 0; i--) {
      for (j = v[i] - 'a' + 1; !is_bigger && j < 26; j++) {
        if (have[j]) {
          is_bigger = 1;
          fill = 'a' + j;
        }
      }
    }
    for (j = 0; j < i + 1; j++) {
      putchar(v[j]);
    }
    putchar(fill);
    for (i = 0; i < 26; i++) {
      if (have[i]) {
        fill = 'a' + i;
        break;
      }
    }
    for (j++; j < k; j++) {
      putchar(fill);
    }
  }
  return 0;
}

```