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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

const int mul[2] = {233, 19260817};
const int mod[2] = {998244353, 1000000007};

#define pb push_back

int n;
vector <int> a, b, na, nb, pw[2], hsh[2];
int get(int l, int r, int k) {return (hsh[k][r] + mod[k] - 1LL * pw[k][r - l] * hsh[k][l] % mod[k]) % mod[k];}

int main() {
	ios :: sync_with_stdio(0);
	cin >> n;
	for (int i = 0; i < n; ++ i) {int x; cin >> x; a.pb(x);}
	for (int i = 0; i < n; ++ i) {int x; cin >> x; b.pb(x);}
	for (int i = 0; i < n; ++ i) na.pb(a[i] ^ a[(i + 1) % n]);
	for (int i = 0; i < n; ++ i) na.pb(na[i]);
	for (int i = 0; i < n; ++ i) nb.pb(b[i] ^ b[(i + 1) % n]);
	int hb[2];
	for (int k = 0; k < 2; ++ k) {
		pw[k].pb(1); for (int i = 1; i <= n * 2; ++ i) pw[k].pb(1LL * mul[k] * pw[k].back() % mod[k]);
		hsh[k].pb(0); for (int i = 0; i < n * 2; ++ i) hsh[k].pb((1LL * mul[k] * hsh[k].back() + na[i]) % mod[k]);
		hb[k] = 0; for (int i = 0; i < n - 1; ++ i) hb[k] = (1LL * mul[k] * hb[k] + nb[i]) % mod[k];
	}
	for (int i = 0; i < n; ++ i)
		if (get(i, i + n - 1, 0) == hb[0] && get(i, i + n - 1, 1) == hb[1] && na[i + n - 1] == nb[n - 1])
			cout << i << " " << (a[i] ^ b[0]) << "\n";
	return 0;
}