#include <bits/stdc++.h>
using namespace std;
const int N = 1 << 17;
int n, k, jhghajkhja, oauighgajk;
char a[N], b[N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int i;
  scanf("%d %d", &n, &k);
  scanf("%s", a + 1);
  for (i = 1; i <= n; i++) b[i] = a[i];
  for (i = 1; i <= n && k > 0; i++) {
    jhghajkhja = a[i] - 'a';
    oauighgajk = 'z' - a[i];
    if (jhghajkhja > oauighgajk) {
      jhghajkhja = min(jhghajkhja, k);
      k -= jhghajkhja;
      b[i] -= jhghajkhja;
    } else {
      oauighgajk = min(oauighgajk, k);
      k -= oauighgajk;
      b[i] += oauighgajk;
    }
  }
  if (k == 0) {
    for (i = 1; i <= n; i++) printf("%c", b[i]);
    printf("\n");
  } else {
    printf("-1\n");
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
const int N = 1 << 17;
int n, k, jhghajkhja, oauighgajk;
char a[N], b[N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int i;
  scanf("%d %d", &n, &k);
  scanf("%s", a + 1);
  for (i = 1; i <= n; i++) b[i] = a[i];
  for (i = 1; i <= n && k > 0; i++) {
    jhghajkhja = a[i] - 'a';
    oauighgajk = 'z' - a[i];
    if (jhghajkhja > oauighgajk) {
      jhghajkhja = min(jhghajkhja, k);
      k -= jhghajkhja;
      b[i] -= jhghajkhja;
    } else {
      oauighgajk = min(oauighgajk, k);
      k -= oauighgajk;
      b[i] += oauighgajk;
    }
  }
  if (k == 0) {
    for (i = 1; i <= n; i++) printf("%c", b[i]);
    printf("\n");
  } else {
    printf("-1\n");
  }
  return 0;
}

```