Choose the incorrect statement regarding time complexity of union$-$find operation:

Union$-$by$-$size and Union$-$by$-$rank both improve the time complexity to O$($M log$($N$))$ for a sequence of M operations and N elements

To finish the entire equivalence class computation algorithm, we need to go over each pair of elements, so if we use union$-$by$-$rank with path compression for find operation, then the overall time complexity is O$($N$^2$ log$*$N$),$ where log$*$N denotes the inverse Ackermann function

When we use arbitrary union and simple find for union$-$find operation, the worst$-$case time complexity is O$($MN$)$ for a sequence of M operations and N elements

Inverse Ackermann function does not depend on N and is a constant factor