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.*;
import java.util.*;
public class lol
{
	BufferedReader in;
	StringTokenizer st;
	PrintWriter ob;
	public static void main(String[] args) throws IOException {
		new lol().runner();
	}
	void runner() throws IOException {
		in=new BufferedReader(new InputStreamReader(System.in));
		ob=new PrintWriter(System.out);
		solver();
		ob.flush();
	}
	void solver() throws IOException {
		char s[]=ns().toCharArray();
		if(false) {
			System.out.println("hie");
		}
		int len=s.length;
		if(false) {
			System.out.println("hie");
		}
		if(s[0]=='a') {
			if(false) {
			System.out.println("hie");
		}
			int k=0;
			for(int i=1;i<len;i++) {
		if(false) {
			System.out.println("hie");
		}
				if(s[i]!='a') {
					k=1;
		if(false) {
			System.out.println("hie");
		}
					s[i]=(char)(s[i]-1);
				}
				else if(s[i]=='a' && k==1) {
		if(false) {
			System.out.println("hie");
		}
					break;
				}
			}
		if(false) {
			System.out.println("hie");
		}
			if(k==0)
				s[len-1]='z';
		}
		else {
			for(int i=0;i<len;i++) {
		if(false) {
			System.out.println("hie");
		}
				if(s[i]=='a')
					break;
				else {
		if(false) {
			System.out.println("hie");
		}
					s[i]=(char)(s[i]-1);
				}
			}
		}
		if(false) {
			System.out.println("hie");
		}
		ob.println(s);
	}
	String ns() throws IOException {
		return nextToken();
	}
	long nl() throws IOException {
		return Long.parseLong(nextToken());
	}
	int ni() throws IOException {
		return Integer.parseInt(nextToken());
	}
	double nd() throws IOException {
		return Double.parseDouble(nextToken());
	} 
	String nextToken() throws IOException {
		if(st==null || !st.hasMoreTokens())
			st=new StringTokenizer(in.readLine());
		return st.nextToken();
	}
}