Mr. Takahashi has a string s consisting of lowercase English letters. He repeats the following operation on s exactly K times.

* Choose an arbitrary letter on s and change that letter to the next alphabet. Note that the next letter of `z` is `a`.



For example, if you perform an operation for the second letter on `aaz`, `aaz` becomes `abz`. If you then perform an operation for the third letter on `abz`, `abz` becomes `aba`.

Mr. Takahashi wants to have the lexicographically smallest string after performing exactly K operations on s. Find the such string.

Constraints

* 1≤|s|≤10^5
* All letters in s are lowercase English letters.
* 1≤K≤10^9

Input

The input is given from Standard Input in the following format:


s
K


Output

Print the lexicographically smallest string after performing exactly K operations on s.

Examples

Input

xyz
4


Output

aya


Input

a
25


Output

z


Input

codefestival
100


Output

aaaafeaaivap
import java.io.InputStream;
import java.io.IOException;
import java.io.PrintStream;
import java.util.*;

public class Main {

  public static void main(String[] args) {
    solve(System.in, System.out);
  }

  static void solve(InputStream is, PrintStream os) {
    Scanner sc = new Scanner(is);

    /* read */
    String s = sc.next();
    int k = sc.nextInt();
    StringBuilder b = new StringBuilder();
    for (int i = 0; i < s.length() - 1; i++) {
      char c = s.charAt(i);
      if (c == 'a') {
        b.append('a');
      } else {
        int need = 'z' + 1 - c;
        if (need <= k) {
          k -= need;
          b.append('a');
        } else {
          b.append(c);
        }
      }
    }
    char c = s.charAt(s.length() - 1);
    k = k % 26;
    c += k;
    if ('z' < c) c -= 26;
    b.append(c);
    os.println(b.toString());
  }

  private static class Scanner {

    private final InputStream is;
    private final byte[] buffer = new byte[1024];
    private int ptr = 0;
    private int buflen = 0;

    Scanner(InputStream is) {
      this.is = is;
    }

    private boolean hasNextByte() {
      if (ptr < buflen) {
        return true;
      } else {
        ptr = 0;
        try {
          buflen = is.read(buffer);
        } catch (IOException e) {
          e.printStackTrace();
        }
        if (buflen <= 0) {
          return false;
        }
      }
      return true;
    }

    private int readByte() {
      if (hasNextByte()) {
        return buffer[ptr++];
      } else {
        return -1;
      }
    }

    private static boolean isPrintableChar(int c) {
      return 33 <= c && c <= 126;
    }

    public boolean hasNext() {
      while (hasNextByte() && !isPrintableChar(buffer[ptr])) {
        ptr++;
      }
      return hasNextByte();
    }

    public String next() {
      if (!hasNext()) {
        throw new NoSuchElementException();
      }
      StringBuilder sb = new StringBuilder();
      int b = readByte();
      while (isPrintableChar(b)) {
        sb.appendCodePoint(b);
        b = readByte();
      }
      return sb.toString();
    }

    public long nextLong() {
      if (!hasNext()) {
        throw new NoSuchElementException();
      }
      long n = 0;
      boolean minus = false;
      int b = readByte();
      if (b == '-') {
        minus = true;
        b = readByte();
      }
      if (b < '0' || '9' < b) {
        throw new NumberFormatException();
      }
      while (true) {
        if ('0' <= b && b <= '9') {
          n *= 10;
          n += b - '0';
        } else if (b == -1 || !isPrintableChar(b)) {
          return minus ? -n : n;
        } else {
          throw new NumberFormatException();
        }
        b = readByte();
      }
    }

    public int nextInt() {
      long nl = nextLong();
      if (nl < Integer.MIN_VALUE || nl > Integer.MAX_VALUE) {
        throw new NumberFormatException();
      }
      return (int) nl;
    }

    public double nextDouble() {
      return Double.parseDouble(next());
    }
  }
}
