Do it in a minute

Following are the first 10 terms of the sequence of the 2016-digit numbers with all their digits repeating in ascending order;

$0123456789012345678901234567890 \dots 123456789012345$ $123456789012345678901234567890 \dots 1234567890123456$ $23456789012345678901234567890 \dots 12345678901234567$ $3456789012345678901234567890 \dots 123456789012345678$ $456789012345678901234567890 \dots 1234567890123456789$ $56789012345678901234567890 \dots 12345678901234567890$ $6789012345678901234567890 \dots 123456789012345678901$ $789012345678901234567890 \dots 1234567890123456789012$ $89012345678901234567890 \dots 12345678901234567890123$ $9012345678901234567890 \dots 123456789012345678901234$

What are the last $3$ digits of the $2292016^\text{th}$ term of the sequence?

Clarifications:

• Considered the first one as a 2016-digit number as well, the number has one leading zero and then the pattern continues.
• Each of the numbers has a repeating part $\dots 0123456789 \dots$

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Try Part 2 of this problem as well.