Intro
Beauty lies in the eye of the beholder. Output lies in the choice of the compiler. There are some codes that give different outputs based on what language they are executed in. Take for instance, the code given below:
# include <stdio.h>
# define print(a) int main(){printf("C is better"); return 0;}
print("Python is better")
When executed in C, it prints "C is better". When using a python interpreter, it prints "Python is better".
Challenge
The challenge is a modification of the FizzBuzz challenge. Write a code that gives different outputs based on language it is executed in. When executed with the first language, it prints all numbers from 1 to 1000 (both inclusive) which are not divisible by 2. If a number is divisible by 2, it outputs "FizzBuzz". When executed with the second language, if a number is not divisible by 3, it is printed. Else, the string "FizzBuzz" is printed.
Example
Output when executed in language 1 would be
1 FizzBuzz 3 FizzBuzz 5 FizzBuzz 7 FizzBuzz 9 FizzBuzz ... (upto FizzBuzz 999 FizzBuzz)
Output when executed in language 2 would be
1 2 FizzBuzz 4 5 FizzBuzz 7 8 FizzBuzz 10 11 FizzBuzz 13 14 FizzBuzz ... (upto 998 FizzBuzz 1000)
Optional Challenge
You can optionally allow the program to execute in more than 2 languages. For the ith language, every multiple of (i+1) is substituted with FizzBuzz. It isn't necessary, but at least 2 languages are compulsory.
Constraints
Need to write a fully functioning code. For instance, a method/function/procedure alone (which could not independently execute) would not be acceptable
Can use only 1 file
All outputs to be printed to standard output (not to standard error)
All other standard rules of code-golf apply
EDIT: Fixed a loophole:
No taking of user inputs during execution
Edit I got a comment saying the question wasn't clear if a non-empty separator is mandatory between the numbers. Assume it's not mandatory
Scoring
Total score = Number of bytes in the program.
Winning
Consider different participation brackets (based on number of languages being used). Person with least characters in each bracket can be considered a winner.