#include <bits/stdc++.h>
using namespace std;
char input[100001];
char output[100001];
int main() {
  int k;
  int len;
  scanf("%d", &len);
  scanf("%d", &k);
  scanf("%s", input);
  int curDistance = 0;
  int totDistance = 0;
  for (int i = 0; i < len; i++) {
    curDistance = min(k, max((int)(input[i] - 'a'), (int)('z' - input[i])));
    if ((char)(input[i] + curDistance) <= 'z' &&
        (char)(input[i] + curDistance) >= 'a')
      output[i] = input[i] + curDistance;
    else if ((char)(input[i] - curDistance) >= 'a' &&
             (char)(input[i] - curDistance) <= 'z') {
      output[i] = input[i] - curDistance;
    } else {
      k = -1;
      break;
    }
    k = k - curDistance;
  }
  output[len] = '\0';
  if (k == 0)
    printf("%s", output);
  else
    printf("-1");
  return 0;
}

### Prompt

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

Limak is a little polar bear. He likes nice strings — strings of length n, consisting of lowercase English letters only.

The distance between two letters is defined as the difference between their positions in the alphabet. For example, <image>, and <image>.

Also, the distance between two nice strings is defined as the sum of distances of corresponding letters. For example, <image>, and <image>.

Limak gives you a nice string s and an integer k. He challenges you to find any nice string s' that <image>. Find any s' satisfying the given conditions, or print "-1" if it's impossible to do so.

As input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use gets/scanf/printf instead of getline/cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java.

Input

The first line contains two integers n and k (1 ≤ n ≤ 105, 0 ≤ k ≤ 106).

The second line contains a string s of length n, consisting of lowercase English letters.

Output

If there is no string satisfying the given conditions then print "-1" (without the quotes).

Otherwise, print any nice string s' that <image>.

Examples

Input

4 26
bear


Output

roar

Input

2 7
af


Output

db


Input

3 1000
hey


Output

-1

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
char input[100001];
char output[100001];
int main() {
  int k;
  int len;
  scanf("%d", &len);
  scanf("%d", &k);
  scanf("%s", input);
  int curDistance = 0;
  int totDistance = 0;
  for (int i = 0; i < len; i++) {
    curDistance = min(k, max((int)(input[i] - 'a'), (int)('z' - input[i])));
    if ((char)(input[i] + curDistance) <= 'z' &&
        (char)(input[i] + curDistance) >= 'a')
      output[i] = input[i] + curDistance;
    else if ((char)(input[i] - curDistance) >= 'a' &&
             (char)(input[i] - curDistance) <= 'z') {
      output[i] = input[i] - curDistance;
    } else {
      k = -1;
      break;
    }
    k = k - curDistance;
  }
  output[len] = '\0';
  if (k == 0)
    printf("%s", output);
  else
    printf("-1");
  return 0;
}

```