I did not write `tray.svg`, because the stated requirements have no solution. Here's the proof (verified by brute-forcing all 720 table↔drink permutations: **zero** satisfy every table).

**The dairy squeeze**
- Tables 1 and 4 both require dairy milk.
- The only dairy drinks are the **Vanilla cortado** and the **Decaf cortado** — two drinks for two tables, so both are consumed by tables 1 and 4.
- That leaves Table 2 (decaf) and Table 3 (vanilla) to be served by the one remaining milk drink, the **Decaf oat vanilla latte** (the only other decaf drink and the only other vanilla drink). One drink can't go to two tables. Contradiction.

**The cold/black squeeze**
- The **Iced oat latte** satisfies no table except Table 5 (cold), so it is forced to Table 5.
- The **Filter coffee** satisfies no table except Table 6 (black), so it is forced to Table 6.
- The **Cold brew** is also cold and black, so Tables 5 and 6 are its only possible recipients — but both are already taken. Contradiction.

Either one of these alone makes a bijection impossible, so I can't honestly draw a tray where every cup is labeled with a table it actually satisfies. At best, any arrangement leaves at least one table with the wrong drink (the maximum is 5 of 6 satisfied).

If it would help, tell me which constraint is flexible (e.g. a table will accept a substitution, or one drink may be held back), and I'll draw the tray accordingly. Otherwise, the correct answer is that the six drinks cannot be distributed to satisfy all six requests.