Understanding Organization of the CRAM bits in bitstream .bin file

Question

For an iCE40 1k device, Following is the snippet from the output of the command "iceunpack -vv example.bin" I could not understand why there are 332x144 bits? My understanding is that [1], the CRAM BLOCK[0] starts at the logic tile (1,1), and it should contain:

• 48 logic tiles, each 54x16,
• 14 IO tiles, each 18x16

How the "332 x 144" is calculated?

Where does the IO tile and logic tiles bits are mapped in CRAM BLOCK[0] bits?

e.g., which bits of CRAM BLOCK[0] indicates the bits for logic tile (1,1) and bits for IO tile (0,1)?

Set bank to 0.
Next command at offset 26: 0x01 0x01
CRAM Data [0]: 332 x 144 bits = 47808 bits = 5976 bytes
Next command at offset 6006: 0x11 0x01

[1]. http://www.clifford.at/icestorm/format.html

Thanks.

Answer 1

height=9x16=144 (1 I/O tile and 8 Logic tiles)

Width=18+42+5x54 = 330 (1 I/O tile, 1 ram tile and 5 Logic tiles) plus "two zero bytes" = 332.