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: Wrong Answer)

#include <bits/stdc++.h>
#define ll long long
#define fastIO cout << fixed << setprecision(12), ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr)
using namespace std;

const int N = 2e5 + 9;

ll z[N+N+N+N], a[N+N], b[N];
vector<ll> va, vb;

vector<ll> vec;
void Z() {
	for(auto it: vb)
		vec.push_back(it);
	vec.push_back(-1);
	for(auto it: va)
		vec.push_back(it);
	for (ll i = 1, l = 0, r = 0; i < (int)vec.size(); ++i) {
		if(i <= r)
			z[i] = min(r - l + 1, z[i - l]);
		while(z[i] + i < (int)vec.size() && vec[z[i]] == vec[z[i] + i])
			++z[i];
		if(i + z[i] - 1 > r) {
			l = i;
			r = i + z[i] - 1;
		}
	}
}

int main() {
	fastIO;
#ifdef LOCAL
	freopen("input.in", "rt", stdin);
#endif
	int n;
	cin >> n;
	for (int i = 0; i < n; ++i)
		cin >> a[i];
	for (int i = 0; i < n; ++i)
		cin >> b[i];
	for (int i = 1; i < n; ++i) {
		va.push_back((ll)a[i] ^ a[i - 1]);
		vb.push_back((ll)b[i] ^ b[i - 1]);
	}
  	for(int i = n; i < n + n; ++i) {
      a[i] = a[i - n];
      va.push_back(a[i] ^ a[i - 1]);      
    }
	Z();
	for (int i = n; i < n + n; ++i) {
		int idx = i - n;
		ll x = a[idx] ^ b[0];
		if(z[i] == n - 1)
			cout << idx << ' ' << x << '\n';
	}
	return 0;
}