#include <iostream>
using namespace std;
const int p[] = {1000000007, 1000000009, 1000000021, 1000000033};
const int c[] = {24809635, 839751792, 823818539, 210819658};

long long modexp(int x, long long e, int m) {
    long long ans = 1, p = x % m;
    while (e > 0) {
        if (e % 2 != 0) ans = (ans * p) % m;
        p = (p * p) % m;
        e >>= 1;
    }
    return ans;
}

struct T {
    long long x[4] = {};
    int *a, n, i;
    T(int *a, int n) : a(a), n(n), i(0) {
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < 4; j++)
                x[j] = ((long long)x[j] * c[j] + a[i]) % p[j];
        }
    }
    void shift() {
        for (int j = 0; j < 4; j++) {
            x[j] = (x[j] * c[j] + a[i] - modexp(c[j], n, p[j]) * a[i]) % p[j];
            if (x[j] < 0) x[j] += p[j];
        }
        i++;
    }
    bool operator==(const T &o) {
        for (int j = 0; j < 4; j++) if (x[j] != o.x[j]) return false;
        return true;
    }
};
int main() {
    int n; cin >> n;
    int a[n], b[n], c[n], d[n];
    for (int i = 0; i < n; i++) cin >> a[i];
    for (int i = 0; i < n; i++) cin >> b[i];
    for (int i = 0; i < n; i++) {
        c[i] = a[i]^a[(i+1)%n];
        d[i] = b[i]^b[(i+1)%n];
    }

    T h1(c, n), h2(d, n);
    for (int i = 0; i < n; i++) {
        if (h1 == h2) {
            cout << i << ' ' << (a[i]^b[0]) << endl;
        }
        h1.shift();
    }
}

### Prompt

Construct a CPP code solution to the problem outlined:

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

### Response

```cpp
#include <iostream>
using namespace std;
const int p[] = {1000000007, 1000000009, 1000000021, 1000000033};
const int c[] = {24809635, 839751792, 823818539, 210819658};

long long modexp(int x, long long e, int m) {
    long long ans = 1, p = x % m;
    while (e > 0) {
        if (e % 2 != 0) ans = (ans * p) % m;
        p = (p * p) % m;
        e >>= 1;
    }
    return ans;
}

struct T {
    long long x[4] = {};
    int *a, n, i;
    T(int *a, int n) : a(a), n(n), i(0) {
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < 4; j++)
                x[j] = ((long long)x[j] * c[j] + a[i]) % p[j];
        }
    }
    void shift() {
        for (int j = 0; j < 4; j++) {
            x[j] = (x[j] * c[j] + a[i] - modexp(c[j], n, p[j]) * a[i]) % p[j];
            if (x[j] < 0) x[j] += p[j];
        }
        i++;
    }
    bool operator==(const T &o) {
        for (int j = 0; j < 4; j++) if (x[j] != o.x[j]) return false;
        return true;
    }
};
int main() {
    int n; cin >> n;
    int a[n], b[n], c[n], d[n];
    for (int i = 0; i < n; i++) cin >> a[i];
    for (int i = 0; i < n; i++) cin >> b[i];
    for (int i = 0; i < n; i++) {
        c[i] = a[i]^a[(i+1)%n];
        d[i] = b[i]^b[(i+1)%n];
    }

    T h1(c, n), h2(d, n);
    for (int i = 0; i < n; i++) {
        if (h1 == h2) {
            cout << i << ' ' << (a[i]^b[0]) << endl;
        }
        h1.shift();
    }
}

```