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 9.2.1) (judge verdict: Wrong Answer)

#include <stdio.h>
//#pragma GCC optimize("O2")
long long int a[200000], b[200000];
long long int a_xor[200000], b_xor[200000];
int main(void){
    int n;
    int m = 15;//baseの数
    long long int base[] = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 43, 47, 53};
    long long int mod = 9007199254740997;
    scanf("%d", &n);
    int i, j, k;
    for(i = 0; i < n; i++){
        scanf("%lld", &a[i]);
    }
    for(i = 0; i < n; i++){
        scanf("%lld", &b[i]);
    }
    //b_iとb_i+1のxorを求める, aも同様に求める
    for(i = 0; i < n; i++){
        b_xor[i] = (b[i] ^ b[(i+1)%n]) + 1;//b []が0になるのを防ぐため1を足す
        a_xor[i] = (a[i] ^ a[(i+1)%n]) + 1;
    }
    // for(i = 0; i < n; i++){
    //     printf("a: %d b: %d\n", a_xor[i], b_xor[i]);
    // }

    //初期(k=0)のhash値を求める
    long long int ahash[m], bhash[m];
    for(j = 0; j < m; j++){
        ahash[j] = 0;
        bhash[j] = 0;
        for(i = 0; i < n; i++){
            ahash[j] += a_xor[i];
            ahash[j] %= mod;
            bhash[j] += b_xor[i];
            bhash[j] %= mod;
            if(i == n - 1){
                break;
            }
            ahash[j] *= base[j];
            ahash[j] %= mod;
            bhash[j] *= base[j];
            bhash[j] %= mod;
        }
    }

    //base^(n-1)%mod の値を求める
    long long int tmp[m];
    for(j = 0; j < m; j++){
        tmp[j] = 1;
        for(i = 0; i < n - 1; i++){
            tmp[j] *= base[j];
            tmp[j] %= mod;
        }
    }

    //kでfor文を回す
    for(i = 0; i < n; i++){
        int cnt = 0;
        for(j = 0; j < m; j++){
            if(ahash[j] == bhash[j]){
                cnt++;
            }
        }
        
        if(cnt == m){
            printf("%lld %lld\n", i, b[0]^a[i]);
        }

        //ahashを1つずらす
        for(j = 0; j < m; j++){
            ahash[j] = (ahash[j] - tmp[j]*a_xor[i])*base[j] + a_xor[i];
            while(ahash[j] < 0){
                ahash[j] += mod; 
            }
            ahash[j] %= mod;
        }
        
    }
    return 0;
}