k
||upto(i+1;j)|| upto(i+1;j)[(k-1)] = i+k
i,j:
, k:(d-1). j-i = d-1 
upto(i;j)[k] = i+k
d
k = 0
[i / upto(i+1;j)][k] = i+k
a:T, as:T List, i:. 0 < i i||as|| [a / as][i] = as[(i-1)]
0
| 1 | k||upto(i+1;j)|| | 1 step |
| 2 | upto(i+1;j)[(k-1)] = i+k | 1 step |
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