#include <bits/stdc++.h>

using namespace std;

int main() {
  cin.tie(nullptr); ios::sync_with_stdio(false);

  int N; cin>>N;
  vector<int> A(N), B(N);
  for (auto& a : A) cin>>a;
  for (auto& a : B) cin>>a;
  vector<int> C(N), D(N);
  for (int i = 0; i < N; i++) {
    C[i] = A[i] xor A[(i+1)%N];
  }
  for (int i = 0; i < N; i++) {
    D[i] = B[i] xor B[(i+1)%N];
  }
  vector<int> lcp(N+1);
  lcp[0] = -1;
  int n = 0;
  for (int i = 1; i < N; i++) {
    if (C[i] == C[n]) {
      lcp[i] = lcp[n];
    } else {
      lcp[i] = n;
      n = lcp[n];
      while (n >= 0 and C[i] != C[n])
        n = lcp[n];
    }
    n++;
  }
  lcp[N] = n;

  int i = 0;
  int p = 0;
  int x = B[0] xor A[0];
  vector<pair<int, int>> ans;
  while (i < N*2-1 and p < N) {
    if (D[i%N] == C[i-p]) {
      if (i-p == N-1) {
        ans.emplace_back((N-p)%N, x);
        p = i+1 - lcp[i+1-p];
        x = B[(i+1)%N] xor A[(i+1-p)%N];
      }
    } else {
      p = i - lcp[i-p];
      x = B[(i+1)%N] xor A[(i+1-p)%N];
    }
    i++;
  }
  sort(ans.begin(), ans.end());
  for (auto a : ans)
    cout << a.first << " " << a.second << endl;

  return 0;
}

### Prompt

Generate a CPP solution to the following problem:

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 <bits/stdc++.h>

using namespace std;

int main() {
  cin.tie(nullptr); ios::sync_with_stdio(false);

  int N; cin>>N;
  vector<int> A(N), B(N);
  for (auto& a : A) cin>>a;
  for (auto& a : B) cin>>a;
  vector<int> C(N), D(N);
  for (int i = 0; i < N; i++) {
    C[i] = A[i] xor A[(i+1)%N];
  }
  for (int i = 0; i < N; i++) {
    D[i] = B[i] xor B[(i+1)%N];
  }
  vector<int> lcp(N+1);
  lcp[0] = -1;
  int n = 0;
  for (int i = 1; i < N; i++) {
    if (C[i] == C[n]) {
      lcp[i] = lcp[n];
    } else {
      lcp[i] = n;
      n = lcp[n];
      while (n >= 0 and C[i] != C[n])
        n = lcp[n];
    }
    n++;
  }
  lcp[N] = n;

  int i = 0;
  int p = 0;
  int x = B[0] xor A[0];
  vector<pair<int, int>> ans;
  while (i < N*2-1 and p < N) {
    if (D[i%N] == C[i-p]) {
      if (i-p == N-1) {
        ans.emplace_back((N-p)%N, x);
        p = i+1 - lcp[i+1-p];
        x = B[(i+1)%N] xor A[(i+1-p)%N];
      }
    } else {
      p = i - lcp[i-p];
      x = B[(i+1)%N] xor A[(i+1-p)%N];
    }
    i++;
  }
  sort(ans.begin(), ans.end());
  for (auto a : ans)
    cout << a.first << " " << a.second << endl;

  return 0;
}

```