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.*;

public final class LettersCyclicShift{

	public static void main(String args[]) throws Exception{

		char[] s = Br.getString().toCharArray();
		int l = s.length;

		int i=0;
		while(i<l && s[i] == 'a')
			i++;

		if(i == l){
			for(int j=0; j<l-1; j++){
				print(s[j]);
			}
			print('z');

			return;
		}

		while(i<l && s[i] != 'a'){
			s[i] = (char)(s[i]-1);
			i++;
		}

		for(int j=0; j<l; j++){
			print(s[j]);
		}
	}

	// class for Reader
	static class Br {

		final static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

		static int getInt() throws IOException{
			return Integer.parseInt(br.readLine());
		}

		static long getLong() throws IOException{
			return Long.parseLong(br.readLine());
		}

		static double getDouble() throws IOException{
			return Double.parseDouble(br.readLine());
		}

		static String getString() throws IOException{
			return br.readLine();
		}

	}

	static void print(char s){
		System.out.print(s);
	}

	static void print(String s){
		System.out.print(s);
	}

	static void println(String s){
		System.out.println(s);
	}
}
