Seek the fastest solution

There are two ulong numbers L1 and L2. You need to set the n-bit (BIT) of L1 to the M-bit (BIT) of L2 ).
For example, L1 is (Binary) 10101000010 ..... (64-bit in total, low-> high), L2 is (Binary) 0101110001 .... (64-bit low-> high in total), set L1 3rd bits (now 1) to L2 1st bits (now 0 ).
I wroteAlgorithmBut it is found that the speed is not fast (used in the encryption algorithm), and most of the time is spent on judgment. I wonder if you can write faster.CodeOr have other optimization skills. The following are my algorithms. Let's take a look at them ~~~ We look forward to your answers. Thank you!

Private   Const Ulong_1 As Ulong =   Ctype ( 1 , Ulong) ' All are 1 ulong
Public   Sub setbit () Sub Setbit ( Byref L1 As Ulong, Byval N As   Integer , Byval L2 As Ulong, Byval M As   Integer )
If (L2 And (Ulong_1 < M )) =   0   Then
L1 = L1 And   Not (Ulong_1 < N)
Else
L1 = L1 Or (Ulong_1 < N)
End   If
End sub

After meditation, the problem is solved: Sub setbit () Sub Setbit ( Byref L1 As Ulong, Byval N As   Integer , Byref L2 As Ulong, Byval M As   Integer )
L2 = (L2 > M) And Culng ( & H1)
L1 = L1 And   Not (Culng ( & H1 < N ))
L1 = L1 + (L2 < N)
End sub

Now it takes only 1000000 ms to execute 31.25 times, and the speed has reached the requirement.
The idea is as follows:
The idea is to shift m in L2 to the rightmost, and then take and with 1, so that M is taken out. For L1, I set N to 0, that is, and 111111 .. 0... 111, 111111 .. 0... 111 comes from 1 <n and then returns the inverse. Finally, we can add the value obtained from L1 to L2 and later.