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)])
(void))))
(time (do ([i 0 (+ i 1)]) ([= i 10000])
(do ([j 0 (+ j 1)]) ([= j 1000])
(void))))
cpu time: 110 real time: 106 gc time: 0
cpu time: 109 real time: 111 gc time: 0
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