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 Scala (2.11.7) (judge verdict: Time Limit Exceeded)

import java.util.Scanner

object KMP {
  // i 番目の直前 res(i) 文字が先頭の res(i) 文字と等しい
  def table(w: Array[Int]): Array[Option[Int]] = {
    val n = w.length
    val res = Array.fill[Option[Int]](n+1)(None)
    var i = 2
    for (i <- 2 to n) {
      res(i-1) match {
        case Some(l) => if (w(i-1) == w(l)) res(i) = Some(l+1)
                        else res(i) = None
        case None => if (w(i-1) == w(0)) res(i) = Some(1)
                     else res(i) = None
      }
    }
    res
  }
  def enumerate(s: List[Int], w: List[Int]): List[Int] = {
    val ss = s.toArray
    val ww = w.toArray
    var m = 0
    var i = 0
    var res = List[Int]()
    val t = table(ww)
    while (m + i < ss.length) {
      if (ss(m+i) == ww(i)) {
        i += 1
        if (i == ww.length) {
          res = m :: res
          m = m + i - t(i).getOrElse(0)
          i = t(i).getOrElse(0)
        }
      } else {
        m = m + i - t(i).getOrElse(-1)
        i = t(i).getOrElse(0)
      }
    }
    res.reverse
  }
}

case class Input(as: List[Int], bs: List[Int]) {
  def solved: Output = {
    val n = as.length
    val a = as.toArray
    val cs = List.tabulate(n)(i => a(i) ^ a((i+1)%n))
    val b = bs.toArray
    val ds = List.tabulate(2*n-1)(i => b(i%n) ^ b((i+1)%n))
    val ks = KMP.enumerate(ds, cs).map(k => (n-k)%n).sorted
    Output(ks.map(k => (k, a(k%n)^b(0))))
  }
}

object Input {
  def fromStdin(): Input = {
    val sc = new Scanner(System.in)
    val n = sc.nextInt()
    Input(List.fill(n)(sc.nextInt), List.fill(n)(sc.nextInt))
  }
}

case class Output(kxs: List[(Int, Int)]) {
  override def toString(): String = {
    kxs.foldLeft("")({ case (acc, (k, x)) => s"${acc}${k} ${x}\n" })
  }

  def print() = printf(this.toString())
}

object Main extends App {
  Input.fromStdin().solved.print()
}