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.util.Scanner;

public class Main {

    private static final Scanner SCANNER = new Scanner(System.in);

    public static void main(String[] args) {
        final String originalString = SCANNER.nextLine();
        int countOperations = SCANNER.nextInt();

        char[] characters = originalString.toCharArray();

        for (int indexLetter = 0; indexLetter < characters.length; ++indexLetter) {
            char letter = characters[indexLetter];
            if (indexLetter != characters.length - 1) {
                int requiredOperations = letter == 'a' ? 0 : 'z' - letter + 1;
                if (requiredOperations <= countOperations) {
                    characters[indexLetter] = push(letter, requiredOperations);
                    countOperations -= requiredOperations;
                }
            } else {
                characters[indexLetter] = push(letter, countOperations);
            }
        }
        System.out.println(String.valueOf(characters));
    }

    private static char push(char letter, int countOperations) {
        letter += countOperations % 26;
        return letter > 'z' ? (char) ('a' + letter - 'z' - 1) : letter;
    }
}
