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
#pragma GCC optimize(3, "Ofast", "inline")
#include <bits/stdc++.h>
#define ll long long
#define bit(x) (x >> bt & 1)
using namespace std;

const int maxn = 2e5 + 5;
int n;
int a[maxn], b[maxn << 1];
int nxt[maxn], cnt[maxn << 1];
void getnxt(int bt) {
	memset(nxt, 0, sizeof(nxt));
	for(int i = 2, j = 0; i <= n; ++i) {
		while(j && bit(a[j + 1]) != bit(a[i]))
			j = nxt[j];
		if(bit(a[j + 1]) == bit(a[i])) j++;
		nxt[i] = j;
	}
}
void kmp(int bt) {
	getnxt(bt);
	for(int i = 1, j = 0; i < 2 * n; ++i) {
		while(j && bit(b[i]) != bit(a[j + 1]))
			j = nxt[j];
		if(bit(b[i]) == bit(a[j + 1])) j++;
		if(j == n) {
			cnt[i - n + 1]++;
			j = nxt[j];
		}
	}
	for(int i = 1, j = 0; i < 2 * n; ++i) {
		while(j && bit(b[i]) != !bit(a[j + 1]))
			j = nxt[j];
		if(bit(b[i]) == !bit(a[j + 1])) j++;
		if(j == n) {
			cnt[i - n + 1]++;
			j = nxt[j];
		}
	}
}
int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr), cout.tie(nullptr);
	cin >> n;
	for(int i = 1; i <= n; ++i) cin >> a[i];
	for(int i = 1; i <= n; ++i) cin >> b[i], b[i + n] = b[i];
	for(int i = 0; i < 30; ++i) kmp(i);
	if(cnt[1] == 30) cout << 0 << " " << (b[1] ^ a[1]) << '\n';
	for(int i = n; i > 1; --i)
		if(cnt[i] == 30) cout << n - i + 1 << " " << (b[i] ^ a[1]) << '\n';
	return 0;
}