import java.io.PrintStream;

public class Test2BShort
{
   public static void main(String[] args)
   {
      PrintStream out = System.out;
      
      final String WORD = args[0];
      
      /*
         Just use InversionCount; if you find two letters that are not
         in alphabetic order then the word cannot be made up of letters
         in alphabetic order.
         
         Note that the complexity of this solution is O(n*n).
         An O(n) solution can be written using a single loop; see
         Test2BShorter.java
      */
      boolean isAlphabetical = true;
      for (int i = 0; i < WORD.length() - 1; i++)
      {
         for (int j = i + 1; j < WORD.length(); j++)
         {
            char ci = WORD.charAt(i);
            char cj = WORD.charAt(j);
            if (ci > cj)
            {
               isAlphabetical = false;
            }
         }
      }
      out.println(isAlphabetical);
   }
}