--description--

N×N disks are placed on a square game board. Each disk has a black side and white side.

At each turn, you may choose a disk and flip all the disks in the same row and the same column as this disk: thus $2 × N - 1$ disks are flipped. The game ends when all disks show their white side. The following example shows a game on a 5×5 board.

It can be proven that 3 is the minimal number of turns to finish this game.

The bottom left disk on the $N×N$ board has coordinates (0, 0); the bottom right disk has coordinates ($N - 1$,$0$) and the top left disk has coordinates ($0$,$N - 1$).

Let $C_N$ be the following configuration of a board with $N × N$ disks: A disk at ($x$, $y$) satisfying $N - 1 \le \sqrt{x^2 + y^2} \lt N$, shows its black side; otherwise, it shows its white side. $C_5$ is shown above.

Let $T(N)$ be the minimal number of turns to finish a game starting from configuration $C_N$ or 0 if configuration $C_N$ is unsolvable. We have shown that $T(5) = 3$. You are also given that $T(10) = 29$ and $T(1\,000) = 395\,253$.

Find $\displaystyle \sum_{i = 3}^{31} T(2^i - i)$.

--hints--

crossFlips() should return 467178235146843500.

assert.strictEqual(crossFlips(), 467178235146843500);

--seed--

--seed-contents--

function crossFlips() {

  return true;
}

crossFlips();

--solutions--

// solution required