Thank you, Laurent.

I mind the performance of in-range, but it seems to be no problem.

#lang racket

(time (for ([i (in-range 10000)])
        (for ([j (in-range 1000)])

(time (do ([i 0 (+ i 1)]) ([= i 10000])
        (do ([j 0 (+ j 1)]) ([= j 1000])

cpu time: 110 real time: 106 gc time: 0
cpu time: 109 real time: 111 gc time: 0

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