This sudoku was designed around this idea : what about ferz-2 where quads specify consecutive numbers ?
In B5, the consecutive pairs must be side by side, not a ferz move apart
Similarilly for consecutive triplets in B1 and B9. In B1 neither 34 nor 45 may be a ferz move apart, so 35 are at ferz and 4 is in one of the other cell. Similarilly for {567} in B9.
6 for R3 in R3C123 => not elsewhere in B1
4 for C3 in R456C3 => not elsewhere in B4
6 for C7 in R456C7 => not elsewhere in B6
4 for R7 in R7C789 => not elsewhere in B9
In B5 :
Since R4C4 = {345} => R3C5, R5C35 <> 4
Since R6C6 = {567} => R5C57, R7C5 <> 6
4 for B4 in R46C3 => R5C24 <> {345}
6 for B6 in R46C7 => R5C68 <> {567}
5 for B4 in R4C23,R5C4 => R4C4 <> 5 (either in same row or at ferz)
5 for B6 in R5C7,R6C78 => R6C6 <> 5
Since R4C4 = {34} => R5C5 <> {34}
{34} for B5 in R4C45 (hidden pair)
{67} for B5 in R6C56
Since R4C45 = {34} => R6C3 = 4
Since R4C4 = {34} => R5C3 <> 3, R6C2 = 3
R4C7 = 6, R6C2 = 7
3 for B1 in R12C1, 3 for B7 in R89C3
{34} for R3 : Since R4C45 = {34} => R456 <> {34} (at ferz of one of the cell in R4C45 : green for pink, blue for orange)
{34} for R3 in R3C789 in B3
3 for D/ => R3C7 = 3 (Symetrically, R7C3 = 7)
4 for R3 in R3C89 (in B3)
Tricky steps !
In B1, quad {345} has consecutive values
Since {34} cannot be at ferz, they must be side by side, dito for {45}
They are placed in an L shape, either orientation, with {35} at ferz and 4 in one of the two other cells
More importantly, {34} cannot go both in R1C2,R2C1 (in red)
Therefore one (and only one) of {34} must be in R1C1 or R2C2 (in blue)
Which is on D\
Since R4C4 = {34}, these 3 cells (R1C1, R2C2, R4C4 in blue) forms a virtual nacked pair on {34} which cannot go elsewhere on D\, R7C7 <> 4
Note : this is similar to "vitual" nacked pair on killer : e.g. one cell with {12} and a cage of 2 cells summing to 5
Since 4 for B3 in R3C89 and 4 for B9 in R7C89 (in green) => R5C7 = 4 (hidden single in C7)
Symetrically for{67} => R5C3 = 6
Since R5C46 = [64] => R4C2,R6C8 <> 5, quads with 5 in B47 => R4C3 = 5, R6C7 = 5, R5C5 = 5
=> R3C1 = 6, R7C2 = 6, R7C9 = 4, R3C8 = 4
From here on, nothing really difficult. Here is a possible path