Problem:
Score : 400 points

Problem Statement

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.

Sample Input 1

xyz
4

Sample Output 1

aya

For example, you can perform the following operations: xyz, yyz, zyz, ayz, aya.

Sample Input 2

a
25

Sample Output 2

z

You have to perform exactly K operations.

Sample Input 3

codefestival
100

Sample Output 3

aaaafeaaivap

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;
import java.io.BufferedReader;
import java.io.InputStreamReader;

public class Main {
    public static void main(String[] args) throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));


		// 1行目数字、2行目文字
		String s = br.readLine();
		long lCount = Integer.parseInt(br.readLine());

		int iLen = s.length();

		StringBuilder sb = new StringBuilder();
		sb.append(s);
					//System.out.println(sb);
		
		char c;
		int iCharNo;
		int i;
		// 先頭から順に各文字を、'a'に変える
		for(i = 0; i<iLen; i++){
			c = sb.charAt(i);
			// a以外
			if (c > 'a') {
				iCharNo = Character.getNumericValue(c);
				//aに変える回数が残ってるなら、変える
				if(lCount >= (36 - iCharNo)){
					lCount = lCount - (36 - iCharNo);
					sb.setCharAt(i, 'a');
				}
			}
		}
			//	System.out.println("sb=[" + sb + "]");
			//	System.out.println("lCount=[" + lCount + "]");
		
		
		//26以上残っているなら、26を一単位として消す
		int iCount = (int)(lCount % 26);
				//System.out.println("iCount=[" + iCount + "]");
		
		// 残った回数を使い切る
		if (iCount > 0) {
			int n = iLen - 1;
			c = sb.charAt(n);
						//System.out.println("c=[" + c + "]");
			
			int k = Character.getNumericValue(c); // 10-35
			k = k - 10;  // 0-25
						//System.out.println("k=[" + k + "]");
			int j = (k + iCount);
						//System.out.println("j=[" + j + "]");
			j = (j % 26); // 0-25
			j = j + 10;
			char a = Character.forDigit(j, 36);
						System.out.println("a=[" + a + "]");
			
			sb.setCharAt(n, a);
		}
		
		System.out.println(sb);

    }
}