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.PrintWriter;
import java.util.StringTokenizer;


public class A {

	

	public static void main(String[] args) throws IOException {
		Scanner sc = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);

		String s = sc.next();
		
		int start = -1;
		for (int i = 0; i < s.length(); i++) {
			if(s.charAt(i) != 'a'){
				start = i;
				break;
			}
		}
		
		int end = s.length();
		for (int i = start + 1; i < s.length(); i++) {
			if(s.charAt(i) == 'a') {
				end = i;
				break;
			}
		}
		
		char[] a = s.toCharArray();
		
		if(start == -1) {
			a[s.length() - 1] = 'z';
		}else
			for (int i = start; i < end; i++) {
				a[i]--;
			}
		
		for (int i = 0; i < a.length; i++) {
			out.print(a[i]);
		}
		
		out.println();

		out.flush();
		out.close();
	}




	static class Scanner 
	{
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream System){br = new BufferedReader(new InputStreamReader(System));}
		public String next() throws IOException 
		{
			while (st == null || !st.hasMoreTokens()) 
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}
		public String nextLine()throws IOException{return br.readLine();}
		public int nextInt() throws IOException {return Integer.parseInt(next());}
		public double nextDouble() throws IOException {return Double.parseDouble(next());}
		public char nextChar()throws IOException{return next().charAt(0);}
		public Long nextLong()throws IOException{return Long.parseLong(next());}
		public boolean ready() throws IOException{return br.ready();}
		public void waitForInput(){for(long i = 0; i < 3e9; i++);}
	}



}