Divide twice in a range's inverse, not four times

Which end of the answer each bound comes from is known from the sign of the
fraction before dividing; taking the min and max of four divisions asked the
question twice. A division is the most expensive thing in that function and
it runs per child per axis.

`many` 0.283 ms a frame to 0.278. Small, and strictly less work.

Co-Authored-By: Claude Opus 5 <noreply@anthropic.com>

core/src/ui/holds.rs

@@ -64,11 +64,12 @@ impl Holds {
         let half_rel = REL_SHIFT - 1;
         let lo = ((self.lo.raw() as i64 - px) * 2 - 3) << half_rel;
         let hi = ((self.hi.raw() as i64 - px) * 2 + 3) << half_rel;
-        let (a, b) = (div_toward(lo, rel, true), div_toward(hi, rel, false));
-        let (c, d) = (div_toward(lo, rel, false), div_toward(hi, rel, true));
+        // Dividing by a negative turns the ends around, so which end each
+        // bound comes from is decided before dividing rather than by taking
+        // the min and max of four divisions.
         match rel > 0 {
-            true => Self::raws(a, b),
-            false => Self::raws(d, c),
+            true => Self::raws(div_toward(lo, rel, true), div_toward(hi, rel, false)),
+            false => Self::raws(div_toward(hi, rel, true), div_toward(lo, rel, false)),
         }
     }