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 (1.15.1) (judge verdict: Accepted)

#![allow(unused_imports)]
#![allow(non_snake_case)]
use std::cmp::*;
use std::collections::*;
use std::io::Write;

#[allow(unused_macros)]
macro_rules! debug {
    ($($e:expr),*) => {
        #[cfg(debug_assertions)]
        $({
            let (e, mut err) = (stringify!($e), std::io::stderr());
            writeln!(err, "{} = {:?}", e, $e).unwrap()
        })*
    };
}

struct MP<T: std::cmp::PartialEq> {
    pi: Vec<i32>,
    s: Vec<T>,
}

impl<T: std::cmp::PartialEq> MP<T> {
    fn new(s: Vec<T>) -> MP<T> {
        let mut pi = vec![0; s.len() + 1];
        pi[0] = -1;
        let mut j = -1;
        for i in 0..s.len() {
            while j >= 0 && s[i] != s[j as usize] {
                j = pi[j as usize];
            }
            j += 1;
            pi[i + 1] = j;
        }
        MP { pi: pi, s: s }
    }

    fn matched_indexes(&self, t: &Vec<T>) -> Vec<usize> {
        let mut j = 0;
        let mut res = vec![];
        for i in 0..t.len() {
            while j >= 0 && self.s[j as usize] != t[i] {
                j = self.pi[j as usize];
            }
            j += 1;
            if j as usize == self.s.len() {
                res.push(1 + i - j as usize);
                j = self.pi[j as usize];
            }
        }
        res
    }
}

fn reduce(a: &Vec<u32>) -> Vec<u32> {
    let mut res = vec![0; a.len() - 1];
    for i in 0..a.len() - 1 {
        res[i] = a[i] ^ a[i + 1];
    }
    res
}

fn main() {
    let n = read::<usize>();
    let mut org_a = read_vec::<u32>();
    let org_b = read_vec::<u32>();
    let mut a_clone = org_a.clone();
    org_a.append(&mut a_clone);
    let a = reduce(&org_a);
    let b = reduce(&org_b);

    // debug!(a);
    // debug!(b);

    let mp = MP::new(b);
    let indexes = mp.matched_indexes(&a);
    for k in indexes {
        if k == n {
            break;
        }
        let x = org_a[k] ^ org_b[0];
        println!("{} {}", k, x);
    }
}

fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>()
        .split_whitespace()
        .map(|e| e.parse().ok().unwrap())
        .collect()
}