Integer Literal - Octals

What is the Output of the below Program:

//OcatalSample.java
public class OcatalSample{
    public static void main(String []args){
      int i = 012;
      int j = 034;
      int k = 056;
      int m = 08; 
      System.out.println("i=" +i + ", j="+j+ ", k="+k+ ", m="+m);
    }
}

public class ReverseNumber {
   public static void main(String[] args) {
      int number = 987;
      System.out.println("Reverse of given number is: "+
               getReverseNumber(number));   
   }

   private static int getReverseNumber(int number) {
      int reverseNumber = 0;
      while (number > 0) {
          int reminder = number % 10;
          reverseNumber = reverseNumber * 10 + reminder;
          number = number / 10;
      }
      return reverseNumber;
   }
}

//Output:
//Reverse of given number is: 789

 Output: It won’t compile. It gives below compilation error
             Exception in thread "main" java.lang.Error: Unresolved compilation problem:
                       The literal 08 of type int is out of range
            at OcatalSample.main(OcatalSample.java:7)

Why??
An integer literal may be expressed in decimal (base 10), hexadecimal (base 16), octal (base 8), or binary (base 2).
Octal representation in java(base 8) :
An integer literal prefixed with 0 is treated as octal. It means when an integer literal starts with 0 in Java, it's assumed to be in octal notation. The digits 8 and 9 are illegal in octal—the digits can range only between 0 and 7.

In the above example #line7 gives an error, because as per above line range of octal is 0 to 7 only.

Note: Generally in octal representation, 7 + 1 is 10.