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.*;
import java.math.*;

public class Solve3 {

  public static void main(String[] args) {
	MyScanner scanner = new MyScanner();
	PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
    
	String s = scanner.nextLine();
	StringBuilder sb = new StringBuilder();
	
	boolean started = false;
	boolean ended = false;
	int changes = 0;
	for(int i = 0; i < s.length(); i++) {
	  char c = s.charAt(i);
	  if(ended) {
	    sb.append(c);
	  } else if(started) {
	    if(c != 'a') {
		  sb.append((char)(c - 1));
		  changes++;
		} else {
		  sb.append(c);
		  ended = true;
		}
	  } else {
	    if(c != 'a') {
		  sb.append((char)(c - 1));
		  changes++;
		  started = true;
		} else {
		  if(changes == 0 && i == s.length() - 1) {
		    sb.append('z');
		  } else {
		    sb.append(c);
		  }
		}
	  }
	}
    
	String res = sb.toString();
	
	out.println(res);
	out.close();
  }

  public static class MyScanner {
      BufferedReader br;
      StringTokenizer st;
 
      public MyScanner() {
         br = new BufferedReader(new InputStreamReader(System.in));
      }
 
      String next() {
          while (st == null || !st.hasMoreElements()) {
              try {
                  st = new StringTokenizer(br.readLine());
              } catch (IOException e) {
                  e.printStackTrace();
              }
          }
          return st.nextToken();
      }
 
      int nextInt() {
          return Integer.parseInt(next());
      }
 
      long nextLong() {
          return Long.parseLong(next());
      }
 
      double nextDouble() {
          return Double.parseDouble(next());
      }
 
      String nextLine(){
          String str = "";
	  try {
	     str = br.readLine();
	  } catch (IOException e) {
	     e.printStackTrace();
	  }
	  return str;
      }

   }
}