#include <iostream>
#include <string>
#include <vector>
#include <set>
#include <queue>
#include <map>
#include <algorithm>
#include <cassert>
#define rep(i,n) for(int i = 0; i < n; ++i )
using namespace std;
using ll = long long;
using P = pair<int,int>;

// Morris-Pratt
template<typename T> struct MP {
  T t; vector<int> a;
  int operator[](int i) { return a[i];}
  MP(const T& t): t(t) {
    a = vector<int>(t.size()+1);
    a[0] = -1;
    
    int j = -1;
    for (int i = 0; i < t.size(); ++i) {
      while (j != -1 && t[j] != t[i]) j = a[j];
      a[i+1] = ++j;
    }
  }
  vector<int> findAll(const T& s) {
    vector<int> res;
    int j = 0;
    for (int i = 0; i < s.size(); ++i) {
      while (j != -1 && t[j] != s[i]) j = a[j];
      if (++j == t.size()) {
        res.push_back(i-t.size()+1);
        j = a[j];
      }
    }
    return res;
  }
};

int main() {
  int n;
  cin >> n;
  vector<ll> a(n+1),b(n+1);
  rep(i,n) cin >> a[i];
  rep(i,n) cin >> b[i];
  a[n] = a[0]; b[n] = b[0];

  vector<int> s(n),t(n);
  rep(i,n){
    s[i] = a[i] ^ a[i+1];
    t[i] = b[i] ^ b[i+1];
  }
  t.insert(t.end(),t.begin(),t.end());

  vector<int> ks = MP<vector<int>>(s).findAll(t);
  rep(i,ks.size()) ks[i] = n-ks[i];
  sort(ks.begin(),ks.end());
  for(int k : ks){
    if (k<n) {
      int x = a[k]^b[0];
      cout << k << " " << x << endl;
    }
  }
}

### Prompt

Your task is to create 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 <iostream>
#include <string>
#include <vector>
#include <set>
#include <queue>
#include <map>
#include <algorithm>
#include <cassert>
#define rep(i,n) for(int i = 0; i < n; ++i )
using namespace std;
using ll = long long;
using P = pair<int,int>;

// Morris-Pratt
template<typename T> struct MP {
  T t; vector<int> a;
  int operator[](int i) { return a[i];}
  MP(const T& t): t(t) {
    a = vector<int>(t.size()+1);
    a[0] = -1;
    
    int j = -1;
    for (int i = 0; i < t.size(); ++i) {
      while (j != -1 && t[j] != t[i]) j = a[j];
      a[i+1] = ++j;
    }
  }
  vector<int> findAll(const T& s) {
    vector<int> res;
    int j = 0;
    for (int i = 0; i < s.size(); ++i) {
      while (j != -1 && t[j] != s[i]) j = a[j];
      if (++j == t.size()) {
        res.push_back(i-t.size()+1);
        j = a[j];
      }
    }
    return res;
  }
};

int main() {
  int n;
  cin >> n;
  vector<ll> a(n+1),b(n+1);
  rep(i,n) cin >> a[i];
  rep(i,n) cin >> b[i];
  a[n] = a[0]; b[n] = b[0];

  vector<int> s(n),t(n);
  rep(i,n){
    s[i] = a[i] ^ a[i+1];
    t[i] = b[i] ^ b[i+1];
  }
  t.insert(t.end(),t.begin(),t.end());

  vector<int> ks = MP<vector<int>>(s).findAll(t);
  rep(i,ks.size()) ks[i] = n-ks[i];
  sort(ks.begin(),ks.end());
  for(int k : ks){
    if (k<n) {
      int x = a[k]^b[0];
      cout << k << " " << x << endl;
    }
  }
}

```