fix: #3621 give range lower precedence than conditional expressions (breaking)

[mvanhorn]

Summary

Range : now has a precedence between assignment and conditional, so the parser matches a strict precedence hierarchy. This targets v16 because it changes how a few expressions parse.

Why this matters

The documented precedence table lists conditional ?: < or < range ::, but the parser didn't follow it: true ? 3 : -1 : 2 : 5 bound the conditional tighter than range, while false or 1:3 bound range tighter than or. That relation is non-transitive, so the parser wasn't strictly precedence-based, and the docs never said so (#3621).

In the issue thread you worked through the options and settled the direction: range can't be higher than conditional (that breaks a ? b : c, since b : c would parse as a range), so range moves below conditional and above assignment. This PR implements that, with the consequences you called out.

Changes

• src/expression/parse.js: reordered the recursive-descent chain to parseAssignment -> parseRange -> parseConditional -> parseLogicalOr -> .... parseConversion now calls parseAddSubtract (range left that slot). Removed the conditionalLevel bookkeeping entirely (all 8 uses) — a : at range level is now unambiguously a range separator because conditional branches parse at a higher level that never consumes :. Net −27 lines in the parser.
• src/expression/operators.js: moved the range band directly above conditional (and below assignment) so toString() / toTex() parenthesization stays consistent with parse precedence and round-tripping remains lossless.
• docs/expressions/syntax.md and HISTORY.md: updated the precedence table and documented the breaking behavior changes.

Accepted behavior changes:

• true ? 3 : -1 : 2 : 5 -> (true ? 3 : -1) : 2 : 5
• a:b==c:d -> a:(b==c):d
• true ? a = 1 : 7 is now a syntax error; use true ? (a=1) : 7
• a = 1:5 still works (range stays above assignment)

Testing

Added #3621 regression tests and updated the tests that encoded the old ordering:

• true ? 3 : -1 : 2 : 5 parses as a RangeNode whose start is true ? 3 : (-1), evaluating to [3, 5]
• false or true ? 1 : 2 still returns 1 (conditional below or)
• false or 1:3 parses as a range with start false or 1 -> [1, 2, 3]
• a:b==c:d -> a:(b == c):d; true ? a = 1 : 7 throws; true ? (a=1) : 7 works; a = 1:5, A[1:3], f(1:3), 1:2:10 unchanged
• toString() round-trip and toTex() parenthesization for the reordered operators

Full expression suite (npx mocha test/unit-tests/expression/**/*.test.js): 2213 passing, no regressions.

Closes #3621