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.lang.*;
import java.util.*;
import java.io.*;


public class Shift {
	
	public static void main(String[] args) throws java.lang.Exception {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		InputReader in = new InputReader(inputStream);
		PrintWriter out = new PrintWriter(outputStream);
		TaskA solver = new TaskA();
		solver.solve(in, out);
		out.close();
	}
}

class TaskA {
	
	public void solve(InputReader in, PrintWriter out) {
		String s = in.next();
		int len = s.length();
		StringBuilder sb = new StringBuilder();
		
		int i = 0;
		
		while (i<len && s.charAt(i)=='a') {
			sb.append('a');
			++i;
		}
		if (i >= len) {
			sb.setCharAt(len-1, 'z');
			out.println(sb);
			return ;
		}
		
		while (i<len && s.charAt(i)!='a') {
			sb.append((char)(s.charAt(i) - 1));
			++i;
		}
		while (i < len) {
			sb.append(s.charAt(i));
			++i;
		}
		
		out.println(sb);
	}
}

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());
	}
	
	public double nextDouble() {
		return Double.parseDouble(next());
	}
}