# [Pyth], 10 bytes <pre><code>VT.dC"¼€ </code></pre> [Try it online!][TIO-k41mt12z] [Pyth]: https://github.com/isaacg1/pyth [TIO-k41mt12z]: https://tio.run/##K6gsyfj/PyxEL8VZ6dAeSeFDDf//AwA "Pyth – Try It Online" There are [exactly 31,557,600 seconds in the Julian astronomical year.](https://en.wikipedia.org/wiki/Year), totalling 31,557,600,000 seconds in 1000 years. 2^32 is about a tenth of this, so we just wait for 3,155,760,000 seconds ten times. Note that <code>€</code> is a blank codepoint in the TIO, not sure why it translates to this on SE [Here](https://tio.run/##K6gsyfj/31np0B5J4UMNSv//AwA "Pyth – Try It Online") you can see that <code>C"¼€ </code>is equal to 3,155,760,000 [And here](https://tio.run/##K6gsyfj/P8xIL8VZiVvp/38A "Pyth – Try It Online") is an example that waits for only 22 seconds using a similar method # [Pyth], 8 bytes <pre><code>.dC"XúÃ </code></pre> [Try it online!][TIO-k41mx8pl] [Pyth]: https://github.com/isaacg1/pyth [TIO-k41mx8pl]: https://tio.run/##K6gsyfj/Xy/FWYk94vCuw83//wMA "Pyth – Try It Online" Alternatively, this one just uses <code>C"XúÃ</code> for 31,557,600,000. I thought it was more in line with the spirit of the challenge to have a 2^32 limit, though