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.IOException;
import java.io.InputStream;
import java.io.PrintStream;
import java.util.NoSuchElementException;

public class Main {
	static FastScanner sc = new FastScanner();
	static PrintStream out = System.out;

	public static void main(String args[]) {
		String s = next();
		int K = ni();
		int len = s.length();

		for(int i=0;i<len-1;i++) {
			char c = s.charAt(i);
			int a = 'z' + 1 - c;
			if (a <= K && c != 'a') {
				K -= a;
				System.out.printf("%c", 'a');
			} else {
				System.out.printf("%c", c);
			}
		}
		
		K %= 26;
		int c = s.charAt(s.length()-1);
		c += K;
		c -= 'a';
		c %= 26;
		c += 'a';
		System.out.printf("%c\n", c);
	}

	private static String next() {
		return sc.next();
	}

	private static int ni() {
		return sc.nextInt();
	}

	private static long nl() {
		return sc.nextLong();
	}

	private static int[] na(int n) {
		int[] a = new int[n];
		for(int i = 0;i < n;i++) a[i] = ni();
		return a;
	}

	private static long[] nal(int n) {
		long[] a = new long[n];
		for(int i = 0;i < n;i++) a[i] = ni();
		return a;
	}

    private static double nd() {
        return Double.parseDouble( next() );
    }
}

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 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();
        }
    }
    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();
        }
    }
}
