Souvlakity should give Kalamakity one of the 1s in the middle of the first and last 1s

When there are 2 ones, Souvlakity doesn't have a middle 1, so has to give Kalamakity one of the 1s, then Kalamakity can give the position next to the remaining 1.

So its $$$1$$$ when the count of 1s is 2

If the count of 1s is 2, answer is 1

Else its the last minus first 1 position

We only need to care about the min and max of the array

The answer is max($$$x$$$ OR $$$y$$$, $$$(x + 1)$$$ OR $$$y$$$, $$$(x + 2)$$$ OR $$$y$$$,... $$$y$$$ OR $$$y$$$)

where $$$x$$$ is the min and $$$y$$$ is the max

Though it'll TLE because $$$a_i \le 1e9$$$. Try to optimize this, consider its binary representations

Read Hint2 first

Consider their binary representations $$$B_x$$$ and $$$B_y$$$

Find the first index where $$$B_x \ne B_y$$$, every index before it is exactly the same as $$$B_x$$$, every index (including this index) onward will be 1

The only values we need to care about is the count of evens and the count of odds

There are only 2 cases where Larpity wins, one of them is already in the sample

Larpity only wins there is exactly 1 even and 1 odd (in the sample)

The other case is when there are no evens and there are exactly 3 odds

Otherwise, Moggity will always force a win

Bonus: Think about how Moggity can force a win

The number of detectors is $$$(n + 1) / 2$$$

Here is the answer for $$$n = 4$$$:

2 1 3 2 4

$$$n = 5$$$:

` 3

1 3

2 4

3 5 `

$$$n = 6$$$:

` 3

1 3

2 4

3 5 `

Try to spot the pattern

First, output $$$(n+1)/2$$$

Then output $$$i$$$ and $$$i + (n-1)/2$$$ for each $$$1 \le i \le (n+1)/2$$$