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: Wrong Answer)

def f(ary)
  n = ary.size
  res = Array.new(n)
  n.times do |i|
    res[i] = ary[i] ^ ary[(i+1)%n]
  end
  res
end
def mp_table(ary)
  res = Array.new(ary.size)
  res[0] = 0
  cnt = 0
  1.upto(ary.size - 1) do |i|
    if ary[cnt] == ary[i]
      cnt += 1
    else
      cnt = 0
    end
    res[i] = cnt
  end
  res.unshift(-1)
end

N = gets.to_i
A = gets.split.map(&:to_i)
B = gets.split.map(&:to_i)

xa = f(A)
xb = f(B)
xa += xa
mp = mp_table(xb)
cnt = 0
ans = []
(2 * N).times do |i|
  if cnt == N
    k = i - N
    ans << [k, A[k] ^ B[0]]
    cnt = mp[cnt]
  end
  
  if xb[cnt] == xa[i]
    cnt += 1
  elsif cnt == 0
    next
  else
    cnt = mp[cnt]
  end
end

puts ans.map{|a| a.join(" ")}.join("\n")