A little puzzle game V2

Try to make a legal position where white just ckeckmate black with the most possible legal amount of points between black and white (black being the 'richer') with 3 rules:
-the black king can't be near a border
-there must be a move that deliver the checkmate so that previous position must be legal too.
-no black or white piece can be placed in a square with number 1, 2 or 3 or with letter e, f, g or h, not even in the previous position of the checkmate (which makes the checker very small)

(I'm sorry to make this a double thread but I thought that making a new with the proper rules would be better)

I'm now pretty sure that there is only one set of solutions and its symetry where the 2 knights, one of the bishop, one rook, both kings, the white pawn and blank spaces are in a mandatory place.
Take my solution here: lichess.org/analysis/qqqq4/qrqq4/nnkq4/K1bb4/1Prq4/8/8/8_w_-_-_0_1?color=white#1
The position can be symetrically inverted with the imaginary axis going between b and c, but if we put this aside, both the knights, the bishop in c5, the rook in c4, both the kings and the blank spaces are mandatory. The light square bishop can then be placed in any white remaining square and the rest of the squares can be filled with queens or the remaining rook.
I've tried to narrow down the number of possible solutions by adding some rule like the last move is the only legal move but that makes the problem non solvable.
If I'm right, that means the number of possibilities are 5 (possible squares for remaining bishop) * 10 (possible square after that for the remaining rook) * 2 (symetry) = 100

The symetry thing can be rid of by changing the rule 1 to "no king can be near a border", making 50 possible solutions ...
I'm surprised this little problem didn't interest anyone. This is fun!

I just found a nice way of narrowing it down to a single solution only. By adding the extra rule 4- all different black pieces must be close one next to each other (both knights, bishop, rooks, all queens must be adjacents). I will make a final version of the problem including this extra rule.