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.BufferedWriter;
import java.io.FileReader;
import java.io.FileWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;

public class Acyclic_Shift {

	public static void main(String[] args)throws IOException {
		
		BufferedReader ob=new BufferedReader(new InputStreamReader(System.in));
		String s=ob.readLine();
		char ch[]=s.toCharArray();
		int n=s.length();
		int index=0;
		if(ch[0]!='a')
		{
			for(int i=0;i<n;i++)
			{
				if(ch[i]!='a')
				{
					ch[i]-=1;
				}
				else 
				{
					break;
				}
			}
		}
		else
		{
			for(int i=0;i<n;i++)
			{
				if(ch[i]=='a')
				{
					index++;
				}
				else
				{
					break;
				}
			}
//			System.out.println("index = "+index);
			if(index==n)
			{
				ch[n-1]='z';
			}
			else
			{
				for(int i=index;i<n;i++)
				{
					if(ch[i]!='a')
					{
						ch[i]-=1;
					}
					else
					{
						break;
					}
				}
			}
			
		}
		for(int i=0;i<n;i++)
		{
			System.out.print(ch[i]);
		}
	}

}
