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>

#define long long long int
using namespace std;

// @author: pashka

int main() {
    ios::sync_with_stdio(false);

    int n;
    cin >> n;
    vector<int> a(n);
    vector<int> b(n);
    for (int i = 0; i < n; i++) {
        cin >> a[i];
    }
    for (int i = 0; i < n; i++) {
        cin >> b[i];
    }

    vector<int> aa(n);
    vector<int> bb(n);
    for (int i = 0; i < n; i++) {
        aa[i] = a[i] ^ (a[(i + 1) % n]);
        bb[i] = b[i] ^ (b[(i + 1) % n]);
    }
    
    vector<int> s;
    for (int i = 0; i < n; i++) {
        s.push_back(bb[i]);
    }
    s.push_back(-1);
    for (int i = 0; i < n; i++) {
        s.push_back(aa[i]);
    }
    for (int i = 0; i < n - 1; i++) {
        s.push_back(aa[i]);
    }
    
    vector<int> p(s.size() + 1);
    p[0] = -1;
    for (int i = 1; i < p.size(); i++) {
        int k = p[i - 1];
        while (k >= 0 && s[i - 1] != s[k]) k = p[k];
        p[i] = k + 1;
        if (p[i] == n) {
            int k = i - n - n - 1;
            int x = b[0] ^ a[k];
            cout << k << " " << x << "\n";
        }
    }
    
    return 0;
}