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 C (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>

int main(){
  int N;
  scanf("%d", &N);
  
  int a[N], b[N];
  for(int i = 0; i < N; i++) scanf("%d", a + i);
  for(int i = 0; i < N; i++) scanf("%d", b + i);
  
  int table[N + 1];
  table[0] = -1;
  for(int i = 0, j = -1; i < N; i++){
    while(j >= 0 && (a[i % N] ^ a[(i + 1) % N]) != (a[j % N] ^ a[(j + 1) % N])) j = table[j];
    table[i + 1] = ++j;
  }
  
  int ans[N];
  for(int i = 0; i < N; i++) ans[i] = -1;
  
  for(int i = 0, p = 0; i + p < 2 * N;){
    if((a[p % N] ^ a[(p + 1) % N]) == (b[(i + p) % N] ^ b[(i + p + 1) % N])){
      if(++p == N){
        ans[(N - i) % N] = a[(N - i) % N] ^ b[0];
        i = i + p - table[p];
        p = table[p];
      }
    }else{
      i = i + p - table[p];
      if(p) p = table[p];
    }
  }
  
  for(int i = 0; i < N; i++){
    if(ans[i] != -1){
      printf("%d %d\n", i, ans[i]);
    }
  }
  
  return 0;
}