Problem:
Score : 600 points

Problem Statement

Given are two sequences a=\{a_0,\ldots,a_{N-1}\} and b=\{b_0,\ldots,b_{N-1}\} of N non-negative integers each.

Snuke will choose an integer k such that 0 \leq k < N and an integer x not less than 0, to make a new sequence of length N, a'=\{a_0',\ldots,a_{N-1}'\}, as follows:

a_i'= a_{i+k \mod N}\ XOR \ x

Find all pairs (k,x) such that a' will be equal to b.

What is \mbox{ XOR }?

The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:

When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.

For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)

Constraints

1 \leq N \leq 2 \times 10^5

0 \leq a_i,b_i < 2^{30}

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
a_0 a_1 ... a_{N-1}
b_0 b_1 ... b_{N-1}

Output

Print all pairs (k, x) such that a' and b will be equal, using one line for each pair, in ascending order of k (ascending order of x for pairs with the same k).

If there are no such pairs, the output should be empty.

Sample Input 1

3
0 2 1
1 2 3

Sample Output 1

1 3

If (k,x)=(1,3),

a_0'=(a_1\ XOR \ 3)=1

a_1'=(a_2\ XOR \ 3)=2

a_2'=(a_0\ XOR \ 3)=3

and we have a' = b.

Sample Input 2

5
0 0 0 0 0
2 2 2 2 2

Sample Output 2

0 2
1 2
2 2
3 2
4 2

Sample Input 3

6
0 1 3 7 6 4
1 5 4 6 2 3

Sample Output 3

2 2
5 5

Sample Input 4

2
1 2
0 0

Sample Output 4

No pairs may satisfy the condition.

Solution in Rust (1.42.0) (judge verdict: Time Limit Exceeded)

fn main() {
	let mut s = String::new();
	use std::io::Read;
	std::io::stdin().read_to_string(&mut s).unwrap();
	let mut s = s.split_whitespace();
	let n: usize = s.next().unwrap().parse().unwrap();
	let a: Vec<u32> = s.by_ref().take(n).map(|a| a.parse().unwrap()).collect();
	let b: Vec<u32> = s.by_ref().take(n).map(|a| a.parse().unwrap()).collect();

	if n == 1 {
		println!("0 {}", a[0] ^ b[0]);
		return;
	}

	let adif: Vec<_> = (0..n)
		.map(|i| a[i] ^ a.get(i + 1).unwrap_or(&a[0]))
		.collect();
	let bdif: Vec<_> = (0..n)
		.map(|i| b[i] ^ b.get(i + 1).unwrap_or(&b[0]))
		.collect();
	let c = [bdif, adif.clone(), adif].concat();
	let z = z_algorithm(&c);
	use std::io::Write;
	let out = std::io::stdout();
	let mut out = std::io::BufWriter::new(out.lock());
	for i in n..2 * n {
		if z[i] >= n - 1 {
			writeln!(out, "{} {}", i - n, b[0] ^ a[i - n]).unwrap();
		}
	}
}

pub fn z_algorithm<T>(s: &[T]) -> Vec<usize>
where
	T: Ord,
{
	if s.len() == 0 {
		Vec::new()
	} else {
		let mut r = vec![0; s.len()];
		let mut j = 0;
		for i in 1..s.len() {
			r[i] = if j + r[i] < s.len() {
				0
			} else {
				std::cmp::min(j + r[j] - i, r[i - j])
			};
			while i + r[i] < s.len() && s[r[i]] == s[i + r[i]] {
				r[i] += 1;
			}
			if j + r[j] < i + r[i] {
				j = i;
			}
		}
		r[0] = s.len();
		r
	}
}