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.

Examples

Input

3
0 2 1
1 2 3


Output

1 3


Input

5
0 0 0 0 0
2 2 2 2 2


Output

0 2
1 2
2 2
3 2
4 2


Input

6
0 1 3 7 6 4
1 5 4 6 2 3


Output

2 2
5 5


Input

2
1 2
0 0


Output
#include <stdio.h>
int n;
unsigned int a[200000];
unsigned int b[200000];
unsigned int x[200000];
unsigned int y[200000];
int z[200000];
unsigned long long q[5], p, k[5], r, m, t, v;
int main() {
	p = 1;
	p <<= 61;
	p--;
	scanf("%d", &n);
	for (int i = 0; i < n; i++) {
		scanf("%d", a + i);
	}
	for (int i = 0; i < n; i++) {
		scanf("%d", b + i);
	}
	x[0] = a[0] ^ a[n - 1];
	for (int i = 1; i < n; i++) {
		x[i] = a[i] ^ a[i - 1];
	}
	y[0] = b[0] ^ b[n - 1];
	for (int i = 1; i < n; i++) {
		y[i] =  b[i] ^ b[i - 1];
	}
	q[0] = 2253887;
	q[1] = 2702209;
	q[2] = 5175323;
	q[3] = 4106041;
	q[4] = 3962909;
	k[0] = 3389;
	k[1] = 4099;
	k[2] = 4663;
	k[3] = 9923;
	k[4] = 18443;
	for (int eq = 0; eq < 5; eq++) {
		p = q[eq];
		r = k[eq];
		v = 1;
		m = 0;
		t = 0;
		for (int i = 0; i < n; i++) {
			m *= r;
			m += y[i];
			m %= p;
			t *= r;
			t += x[i];
			t %= p;
			v *= r;
			v %= p;
		}
		v = (1 + p - v) % p;
		for (int i = 0; i < n; i++) {
			if (t == m)z[i]++;
			t *= r;
			t += v * x[i];
			t %= p;
		}
	}
	for (int i = 0; i < n; i++) {
		if (z[i] == 5) {
			printf("%d %u\n", i, b[0] ^ a[i]);
		}
	}
}