Bus mastering - Who is the most prioritary?.js

/*
Description:
When multiple master devices are connected to a single bus (https://en.wikipedia.org/wiki/System_bus), there needs to be an arbitration in order to choose which of them can have access to the bus (and 'talk' with a slave).

We implement here a very simple model of bus mastering. Given n, a number representing the number of masters connected to the bus, and a fixed priority order (the first master has more access priority than the second and so on...), the task is to choose the selected master. In practice, you are given a string input of length n representing the n masters' requests to get access to the bus, and you should return a string representing the masters, showing which (only one) of them was granted access:

The string 1101 means that master 0, master 1 and master 3 have requested
access to the bus. 
Knowing that master 0 has the greatest priority, the output of the function should be: 1000
Examples
* arbitrate("001000101", 9) -> "001000000" 

* arbitrate("000000101", 9) -> "000000100"

* `n` is not mandatory for solving the kata
Notes
The resulting string (char*) should be allocated in the arbitrate function, and will be free'ed in the tests.

n is always greater or equal to 1.
*/
function arbitrate(input, n) { 
    return input.replace(/1/,'m').replace(/[^m]/gi,'0').replace(/[m]/gi,'1')
}