Let us call the exclusive logical sum XOR and denote the XOR of non-negative integers \(x\) and \(y\) by \(x \oplus y\).

There are \(\binom{2N - 2}{N - 1}\) ways to travel from square \((1, 1)\) to \((N, N)\); in this problem, there are at most about \(3.5 × 10^{10}\) ways. Thus, trying all of them does not finish in the time limit.
Instead, we use the trick called meet-in-the-middle. Roughly, we try to optimize the program by corresponding \(S\) objects into the direct product set of two sets of sizes about \(\sqrt{S}\) and process them.

When traveling from square \((1, 1)\) to \((N, N)\), you pass through a unique square on the squares on the anti-diagonal line, that is, a square \((x, y)\) such that \(x + y = N + 1\). For one path, let \(p\) be the XOR on the squares in the path from square \((1, 1)\) to \((x, y)\), and \(q\) be that from square \((x, y)\) to \((N, N)\); then the path satisfies the condition in the problem statement if and only if \(p \oplus a_{x, y} \oplus q = 0\).

Therefore, let \(P\) be the multiset of the XORs of the integers written on each path from square \((1, 1)\) to \((x, y)\), and \(Q\) be the counterpart from \((x, y)\) to \((N, N)\), then the number of paths satisfying the conditions in the problem statement passing through square \((x, y)\) equals the number of pairs \((p, q)\) such that \(p \in P, q \in Q\), and \(p \oplus a_{x, y} \oplus q = 0\). This can be found by sorting or binary search in a total of \(O((|P| + |Q|) \log (|P| + |Q|))\) time.

Since \(|P| = |Q|\), and the sum of \(|P|\) over all squares \((x, y)\) on the anti-diagonal line is \(2^{N - 1}\), the problem has been solved in a total of \(O(N2^{N})\).