For problem statement at 1000-1999/1300-1399/1360-1369/1365/problemE.txt this is a correct solution, but verifier at 1000-1999/1300-1399/1360-1369/1365/verifierE.go ends with All 100 tests passed. can you fix the verifier? ```go
package main

import (
	"fmt"
	"math/bits"
)

func main() {
	var n int
	fmt.Scan(&n)
	a := make([]uint64, n)
	for i := 0; i < n; i++ {
		fmt.Scan(&a[i])
	}

	var best uint64

	// For subsequence of size 1: value is all bits set in that element
	for i := 0; i < n; i++ {
		if a[i] > best {
			best = a[i]
		}
	}

	// For subsequence of size 2: max(1, 2-2) = 1, so bits where at least 1 of 2 elements has it set
	// That's just OR of two elements
	for i := 0; i < n; i++ {
		for j := i + 1; j < n; j++ {
			val := a[i] | a[j]
			if val > best {
				best = val
			}
		}
	}

	// For subsequence of size 3: max(1, 3-2) = 1, so bits where at least 1 of 3 elements has it set
	// That's just OR of three elements
	// But we want to maximize, so OR of any 3 is >= OR of any 2
	// Actually for size k >= 3: max(1, k-2) = k-2
	// For k=3: need at least 1 out of 3 -> OR
	// For k=4: need at least 2 out of 4
	// For k=5: need at least 3 out of 5
	// ...
	// The key insight: optimal subsequence has size at most 3
	// Because for k>=4, we need k-2 elements to have each bit set
	// Adding more elements doesn't help unless they contribute new bits
	// but the threshold increases too.
	// Actually the optimal is always size 1, 2, or 3.
	// For size 3: threshold is 1 (OR of all 3) - same as size 2
	// Wait, for size 3: max(1, 3-2) = max(1,1) = 1
	// For size 4: max(1, 4-2) = 2
	// So size 3 gives OR which is at least as good as size 4's AND-like condition
	// So optimal is at most size 3.

	for i := 0; i < n; i++ {
		for j := i + 1; j < n; j++ {
			for k := j + 1; k < n; k++ {
				val := a[i] | a[j] | a[k]
				if val > best {
					best = val
				}
			}
		}
	}

	// But wait, we need to verify this reasoning more carefully.
	// For k=3: threshold = 1, value = OR of all elements
	// For k=4: threshold = 2, value = bits where at least 2 of 4 have it
	// A bit contributes if at least 2 of 4 elements have it.
	// Could this be larger than the best OR of 3?
	// If we have elements where each individually has many bits but OR of any 3 misses some pattern...
	// Actually OR of 3 elements from the 4 will catch bits that at least 1 of 3 has,
	// which means at least 1 of 4 has it. The k=4 needs at least 2.
	// So k=4 is strictly harder - its value <= OR of any 3 chosen from those 4.
	// Hence optimal is indeed size <= 3.

	_ = bits.Len // just to use the import
	fmt.Println(best)
}
```