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;

class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String s = sc.next();
        int k = sc.nextInt();
        int len = s.length();
        int rotate[] = new int[len];
        for (int i = 0; i < len; i++) {
            rotate[i] = count(s.charAt(i));
        }
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < len; i++) {

            if (k >= rotate[i]) {
                // 回転させる
                k -= rotate[i];
                // s = s.substring(0, i) + 'a' + s.substring(i + 1);
                sb.append("a");
            } else {
                // 回転出来ない
                sb.append(s.charAt(i));
            }
        }
        k %= 'z' - 'a' + 1;
        char c = sb.charAt(len - 1);
        sb.deleteCharAt(len - 1);
        sb.append((char) (c + k));
        // s = s.substring(0, len - 1) + (char) (s.charAt(len - 1) + k);
        System.out.println(new String(sb));

        sc.close();
    }

    static int count(char c) {
        if (c == 'a') {
            return 0;
        }
        return ((int) ('z' - c)) + 1;
    }
}
