I think about the cross of the election, too, so I composed a simple helpmate three-mover.

(Do you suppose I am monomaniac? I will seek advice.)

8/8/3k4/3b4/1bpKps2/3Q4/3r4/8 w (2+7) |
Problem-566 Manolas Emmanuel Original h#3 2.1.1.1.1.1 (two solutions) |

1.Be6 Ke3+ 2.Ke5 Qf1 3.Rd5 Qxf4#

1.Bf7 Kxe4+ 2.Ke6 Qc3 3.Rd7 Qe5#

Orthogonal - Diagonal transformation.

(Note : After Problem-509, 20 + 36 problems were published here (which supposedly have the numbers 510 - 565), so today's is Problem-566).