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)

fn z_algorithm<T: Eq>(s: &[T]) -> Vec<usize> {
    let len = s.len();
    let mut a = vec![0; len];
    a[0] = len;
    let mut i = 1;
    let mut j = 0;
    while i < len {
        while i + j < len && s[j] == s[i + j] {
            j += 1;
        }
        a[i] = j;
        if j == 0 {
            i += 1;
            continue;
        }
        let mut k = 1;
        while i + k < len && k + a[k] < j {
            a[i + k] = a[k];
            k += 1;
        }
        i += k;
        j -= k;
    }
    a
}

//https://qiita.com/tanakh/items/0ba42c7ca36cd29d0ac8 より
macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
    ($($r:tt)*) => {
        let s = {
            use std::io::Read;
            let mut s = String::new();
            std::io::stdin().read_to_string(&mut s).unwrap();
            s
        };
        let mut iter = s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
}

macro_rules! input_inner {
    ($iter:expr) => {};
    ($iter:expr, ) => {};
    ($iter:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($iter, $t);
        input_inner!{$iter $($r)*}
    };
}

macro_rules! read_value {
    ($iter:expr, ( $($t:tt),* )) => {
        ( $(read_value!($iter, $t)),* )
    };
    ($iter:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($iter, $t)).collect::<Vec<_>>()
    };
    ($iter:expr, chars) => {
        read_value!($iter, String).chars().collect::<Vec<char>>()
    };
    ($iter:expr, usize1) => {
        read_value!($iter, usize) - 1
    };
    ($iter:expr, $t:ty) => {
        $iter.next().unwrap().parse::<$t>().expect("Parse error")
    };
}

//

use std::io::Write;

fn run() {
    let out = std::io::stdout();
    let mut out = std::io::BufWriter::new(out.lock());
    input! {
        n: usize,
        a: [usize; n],
        b: [usize; n],
    }
    let mut ans = vec![true; n];
    for k in 0usize..30 {
        let mut dp = vec![false; n];
        for x in 0usize..2 {
            let mut c = vec![];
            for a in b.iter() {
                c.push(x ^ ((a >> k) & 1));
            }
            c.push(2);
            for _ in 0..2 {
                for b in a.iter() {
                    c.push((b >> k) & 1);
                }
            }
            let z = z_algorithm(&c);
            for i in (n + 1)..(n + 1 + n) {
                if z[i] >= n {
                    dp[i - n - 1] = true;
                }
            }
        }
        for i in 0..n {
            ans[i] &= dp[i];
        }
    }
    for k in 0..n {
        if ans[k] {
            let x = a[k] ^ b[0];
            writeln!(out, "{} {}", k, x).ok();
        }
    }
}

fn main() {
    run();
}