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 Ruby (2.3.3) (judge verdict: Time Limit Exceeded)

N = gets.to_i
As, Bs = readlines.map{|l| l.split.map(&:to_i)}

class RollingHash
  def initialize(ary)
    @l = ary.size
    @b, @m = 1007, 10**9 + 7
    @h, @p = Array.new(@l + 1, 0), Array.new(@l + 1, 1)
    @l.times do |i|
      @h[i + 1] = (@h[i]*@b + ary[i]) % @m
      @p[i + 1] = (@p[i]* @b) % @m
    end
  end

  def get(l, r)
    (@h[r] - @h[l]*@p[r - l]) % @m
  end

  def lcp(rh, a, b)
    l, h = 0, @l + 1 - [a, b].max
    while h - l > 1
      m = (l + h)/2
      self.get(a, a + m) == rh.get(b, b + m) ? l = m : h = m
    end
    l
  end
end

ha, hb = [As, Bs].map{|ary| RollingHash.new(ary.cycle.each_cons(2).take(2*N).map{|x, y| x^y})}
(0 ... N).each{|k| puts "#{k} #{As[k]^Bs[0]}" if ha.lcp(hb, k, 0) >= N}