#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, k, i = 0, diff;
  cin >> n >> k;
  string s;
  cin >> s;
  vector<char> num;
  while ((k > 0) || (i < n)) {
    if (abs(s[i] - 'a') > abs('z' - s[i]))
      diff = s[i] - 'a';
    else
      diff = 'z' - s[i];
    if (abs(diff) <= k) {
      k -= abs(diff);
      if (diff == s[i] - 'a')
        num.push_back(char(int(s[i]) - diff));
      else
        num.push_back(char(int(s[i]) + diff));
      i++;
    } else if ((k < diff) && (k != 0)) {
      if (k > 'z' - s[i])
        num.push_back(char(int(s[i]) - k));
      else
        num.push_back(char(int(s[i]) + k));
      k = 0;
      i++;
    } else if ((k < diff) && (k == 0)) {
      num.push_back(s[i]);
      i++;
    }
  }
  if ((k == 0) && (i == n)) {
    for (auto i : num) cout << i;
  } else
    cout << -1;
  return 0;
}

### Prompt

Your challenge is to write 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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, k, i = 0, diff;
  cin >> n >> k;
  string s;
  cin >> s;
  vector<char> num;
  while ((k > 0) || (i < n)) {
    if (abs(s[i] - 'a') > abs('z' - s[i]))
      diff = s[i] - 'a';
    else
      diff = 'z' - s[i];
    if (abs(diff) <= k) {
      k -= abs(diff);
      if (diff == s[i] - 'a')
        num.push_back(char(int(s[i]) - diff));
      else
        num.push_back(char(int(s[i]) + diff));
      i++;
    } else if ((k < diff) && (k != 0)) {
      if (k > 'z' - s[i])
        num.push_back(char(int(s[i]) - k));
      else
        num.push_back(char(int(s[i]) + k));
      k = 0;
      i++;
    } else if ((k < diff) && (k == 0)) {
      num.push_back(s[i]);
      i++;
    }
  }
  if ((k == 0) && (i == n)) {
    for (auto i : num) cout << i;
  } else
    cout << -1;
  return 0;
}

```