Java Exception Handling understanding issue

I'm not able to understand this program. I expect it to output "Hello World", but instead it prints only the "World". I thought that first the try block would execute, printing "Hello" and " ", then afterward when it encounters a 1/0, it would throw an ArithmeticException. The exception would be caught by catch block, then "World" would be printed.

The program is as follows.

 import java.util.*;
 class exception{
     public static void main(String args[]) 
     {
         try
         {
             System.out.println("Hello"+" "+1/0);
         } 
         catch(ArithmeticException e) 
         {
             System.out.println("World");
         }
     }
 }

First "Hello"+" "+1/0 will be evaluated. And then passed as an argument to System.out.println(...). That's why an exception is thrown before System.out.println(...) would have been called.