You are given a non-empty string s consisting of lowercase English letters. You have to pick exactly one non-empty substring of s and shift all its letters 'z' <image> 'y' <image> 'x' <image> 'b' <image> 'a' <image> 'z'. In other words, each character is replaced with the previous character of English alphabet and 'a' is replaced with 'z'.

What is the lexicographically minimum string that can be obtained from s by performing this shift exactly once?

Input

The only line of the input contains the string s (1 ≤ |s| ≤ 100 000) consisting of lowercase English letters.

Output

Print the lexicographically minimum string that can be obtained from s by shifting letters of exactly one non-empty substring.

Examples

Input

codeforces


Output

bncdenqbdr


Input

abacaba


Output

aaacaba

Note

String s is lexicographically smaller than some other string t of the same length if there exists some 1 ≤ i ≤ |s|, such that s1 = t1, s2 = t2, ..., si - 1 = ti - 1, and si < ti.
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Collections;
import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.PriorityQueue;
import java.util.StringTokenizer;
import java.util.concurrent.LinkedBlockingQueue;



public class Problem2 {
 
	public static void main(String[] args) throws InterruptedException {
		// TODO Auto-generated method stub
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		InputReader in = new InputReader(inputStream);
		PrintWriter out = new PrintWriter(outputStream);
		String s = in.next();
		char[] let = new char[26];
		int beg = s.length();
		boolean yes = true;
		for(int i = 0; i < s.length(); ++i)
		{
			let[s.charAt(i) - 'a']++;
			if(s.charAt(i) != 'a' && yes)
			{
				beg = i;
				yes = false;
			}
		}
		if(let[0] == 0)
		{
			for(int i = 0; i < s.length(); ++i)
			{
				out.print(prev(s.charAt(i)));
			}
		}
		else
		{
			int end = 0;
			if(beg <= s.length()-1)
			{
				for(end = beg; end < s.length(); ++end)
				{
					if(s.charAt(end) == 'a')
					{
						break;
					}
				}
			}
			else
			{
				
				out.print(s.substring(0, s.length() - 1));
                out.print('z');
				out.close();
				return;
			}
			
			out.print(s.substring(0, beg));
			for(int i = beg; i < end; ++i)
			{
				out.print(prev(s.charAt(i)));	
			}
			out.print(s.substring(end, s.length()));
			
		}
		
		out.close();

	}
	
	
	static char prev(char cur)
	{
		if(cur-- == 'a')
		{
			return 'z';
		}
		
		return cur;
		
	}

	
	
	static class InputReader {
	    public BufferedReader reader;
	    public StringTokenizer tokenizer;

	    public InputReader(InputStream stream) {
	        reader = new BufferedReader(new InputStreamReader(stream), 32768);
	        tokenizer = null;
	    }

	    public String next() {
	        while (tokenizer == null || !tokenizer.hasMoreTokens()) {
	            try {
	                tokenizer = new StringTokenizer(reader.readLine());
	            } catch (IOException e) {
	                throw new RuntimeException(e);
	            }
	        }
	        return tokenizer.nextToken();
	    }

	    public int nextInt() {
	        return Integer.parseInt(next());
	    }
	    
	    public long nextLong()
	    {
	    	return Long.parseLong(next());
	    }

	}


}