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

public class Main{
  public static int res = 0;
  public static void main(String[] args)throws Exception {
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    //InputReader in = new InputReader(System.in);
    PrintWriter pw = new PrintWriter(new OutputStreamWriter(System.out));
    String s = "", s1 = "";
    s = br.readLine();
    int c = 0, k = 0;//pw.println(s);
    for(int i = 0; i < s.length(); i++)
    {
      if(s.charAt(i) == 'a')
        c++;
    }
    if(c == s.length())
    {
      for(int i = 0; i < c-1; i++)
        pw.print(s.charAt(i));
      pw.println("z");
      
    }
    else
    {
      boolean b = false;
      for(int i = 0; i < s.length(); i++)
      {
        if(s.charAt(i) == 'a' || b == true)
          pw.print(s.charAt(i));
        else
        {
          while(i < s.length() && s.charAt(i) != 'a')
          {
            int v = (int)s.charAt(i) - 1;
            pw.print((char)(v));
            b = true;
            i++;
          }
          i--;
        }
      }
      pw.println();
    }
    pw.close();
  }
}