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_macros)]
macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        let mut next = || { iter.next().unwrap() };
        input_inner!{next, $($r)*}
    };
    ($($r:tt)*) => {
        let stdin = std::io::stdin();
        let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
        let mut next = move || -> String{
            bytes
                .by_ref()
                .map(|r|r.unwrap() as char)
                .skip_while(|c|c.is_whitespace())
                .take_while(|c|!c.is_whitespace())
                .collect()
        };
        input_inner!{next, $($r)*}
    };
}

#[allow(unused_macros)]
macro_rules! input_inner {
    ($next:expr) => {};
    ($next:expr, ) => {};

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

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

#[allow(unused_macros)]
macro_rules! read_value {
    ($next:expr, ( $($t:tt),* )) => {
        ( $(read_value!($next, $t)),* )
    };

    ($next:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    };

    ($next:expr, chars) => {
        read_value!($next, String).chars().collect::<Vec<char>>()
    };

    ($next:expr, bytes) => {
        read_value!($next, String).into_bytes()
    };

    ($next:expr, usize1) => {
        read_value!($next, usize) - 1
    };

    ($next:expr, $t:ty) => {
        $next().parse::<$t>().expect("Parse error")
    };
}

// int *buildFail(char *p) {
//     int m = strlen(p);
//     int *fail = new int[m+1];
//     int j = fail[0] = -1;
//     for (int i = 1; i <= m; ++i) {
//       while (j >= 0 && p[j] != p[i-1]) j = fail[j];
//       fail[i] = ++j;
//     }
//     return fail;
//   }

// int match(char *t, char *p, int *fail) {
//     int n = strlen(t), m = strlen(p);
//     int count = 0;
//     for (int i = 0, k = 0; i < n; ++i) {
//       while (k >= 0 && p[k] != t[i]) k = fail[k];
//       if (++k >= m) {
//         ++count; // match at t[i-m+1 .. i]
//         k = fail[k];
//       }
//     }
//     return count;
//   }

fn kmp_table<T: Eq>(pattern: &[T], m: usize, mut kmp: Vec<usize>) -> Vec<usize> {
    let mut len: usize = 0;
    kmp[0] = 0;
    let mut i: usize = 1;
    while i < m {
        if pattern[i] == pattern[len] {
            len = len + 1;
            kmp[i] = len;
            i = i + 1;
        } else {
            if len != 0 {
                len = kmp[len - 1];
            } else {
                kmp[i] = 0;
                i = i + 1;
            }
        }
    }
    kmp
}

fn kmp<T: Eq>(string: &[T], pattern: &[T]) -> Option<usize> {
    let n = string.len();
    let m = pattern.len();
    let kmp = kmp_table(pattern, m, vec![0usize; m]);
    let mut i: usize = 0;
    let mut j: usize = 0;
    while i < n {
        if pattern[j] == string[i] {
            i = i + 1;
            j = j + 1;
        }
        if j == m {
            return Some(i - j);
        } else if i < n && pattern[j] != string[i] {
            if j != 0 {
                j = kmp[j - 1];
            } else {
                i = i + 1;
            }
        }
    }
    None
}

fn main() {
    input! {
        n: usize,
        a: [i32; n],
        b: [i32; n],
    }

    let mut xa = a.clone();
    let mut xb = b.clone();

    for i in 0..n {
        xa[i] = a[i] ^ a[(i + 1) % n];
        xb[i] = b[i] ^ b[(i + 1) % n];
    }

    let mut xxa = xa.clone();
    for r in xa.iter() {
        xxa.push(*r);
    }

    let ix = kmp(&xxa, &xb);

    if ix == None {
        return;
    }

    let ix = ix.unwrap();

    let mut sol = vec![false; n];

    for span in 1..n + 1 {
        if n % span != 0 {
            continue;
        }

        let rep = n / span;
        let mut ok = true;
        for i in 0..span {
            let c = xa[i];
            for j in 1..rep {
                if c != xa[i + j * span] {
                    ok = false;
                    break;
                }
            }
            if !ok {
                break;
            }
        }

        if !ok {
            continue;
        }

        for i in 0..rep {
            sol[(ix + i * span) % n] = true;
        }
    }

    for i in 0..n {
        if sol[i] {
            let cur = a[i] ^ b[0];
            println!("{} {}", i, cur);
        }
    }
}