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 <bits/stdc++.h>

using namespace std;

int N;
int arr[400005], brr[400005];
int lps[400005];
bool ok[35][400005];

int main(){
	cin.sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	cin >> N;	
	for(int i = 1; i<=N; i++){
		cin >> arr[i];
	}
	for(int i = 1; i<=N; i++){
		cin >> brr[i];
		brr[i+N] = brr[i];
	}
	lps[1] = 1;
	for(int b = 0; b<30; b++){
		for(int k = 0; k<=1; k++){
			for(int i = 2, j = 1; i<=N; ){
				if((arr[i]&(1<<b)) == (arr[j]&(1<<b))){
					lps[i++] = ++j;
				}
				else if(j-1){
					j = lps[j-1];
				}
				else{
					lps[i++] = j;
				}	
			}
			for(int i = 1, j = 1; i<2*N; ){
				if((brr[i]&(1<<b)) == ((arr[j]&(1<<b))^(k<<b))){
					i++, j++;
					if(j == N+1){
						j = lps[N];
						//cout << b << " " << (i-N) << endl;
						ok[b][i-N] = 1;
					}
				}
				else if(j-1){
					j = lps[j-1];
				}
				else{
					i++;
				}
			}
		}
	}
	vector<pair<int, int>> v;
	for(int i = 1; i<=N; i++){
		int cnt = 0;
		for(int b = 0; b<30; b++){
			cnt += ok[b][i];
		}
		if(cnt == 30){
			v.push_back({(N-i+1)%N, (arr[1]^brr[i])});
		}
	}
	sort(v.begin(), v.end());
	for(auto p : v){
		cout << p.first << " " << p.second << "\n";
	}
}

/*

*/