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: Accepted)


import java.io.IOException;
import java.io.InputStream;
import java.util.NoSuchElementException;
public class Main {
	static final long C =  1000000007;
	static final int CY = 1000000000;
	StringBuilder sb;
	public void calc() {
		sb = new StringBuilder();


		FastScanner sc = new FastScanner();

		char[] s = sc.next().toCharArray();
		long k = sc.nextLong();

		String ans = "";
		//先頭からaにできるやつはしていく
		for (int i=0; i < s.length; i++) {
			int t = 26 - (s[i] - 'a');
			if (t <= k && t != 26) {
				k -= t;
				s[i] = 'a';
			}
		}
		//余ったら最後のやつに全投入
		s[s.length-1] = (char)(((s[s.length-1]-'a'+k)%26)+'a');

		System.out.println(String.valueOf(s));
	}



	public static void main(String[] args) {
		Main main = new Main();
		main.calc();

	}
	static class FastScanner {
		private final InputStream in = System.in;
		private final byte[] buffer = new byte[1024];
		private int ptr = 0;
		private int buflen = 0;
		private boolean hasNextByte() {
			if (ptr < buflen) {
				return true;
			}else{
				ptr = 0;
				try {
					buflen = in.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;}
		private void skipUnprintable() { while(hasNextByte() && !isPrintableChar(buffer[ptr])) ptr++;}
		public boolean hasNext() { skipUnprintable(); 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() {
			if (!hasNext()) throw new NoSuchElementException();
			int 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();
			}
		}
	}
}