#include <bits/stdc++.h>

using namespace std;

#define REP(i,n) for(int i = 0; i < (int)(n); i++)

vector<int> solve(vector<int> a, vector<int> b, int n){
  REP(i, n){
    b.push_back(b[i]);
  }
  a.push_back(2);
  vector<int> e;
  vector<int> t(n + 1);
  t[0] = -1, t[1] = 0;
  int p = 2;
  int q = 0;
  while(p < n + 1){
    if(a[p - 1] == a[q]){
      t[p++] = ++q;
    }else if(q > 0){
      q = t[q];
    }else{
      t[p++] = 0;
    }
  }
  int m = 0;
  p = 0;
  while(m + p < 2 * n){
    if(a[p] == b[m + p]){
      ++p;
    }else{
      m += p - t[p];
      if(p > 0){
        p = t[p];
      }
    }
    if(p == n){
      e.push_back(m);
    }
  }
  return e;
}

int main(){
  int n;
  cin >> n;
  vector<int> a(n), b(n);
  REP(i, n) cin >> a[i];
  REP(i, n) cin >> b[i];
  vector<int> A(n), B(n);
  vector<bool> r0[30];
  vector<bool> r1[30];
  REP(p, 30){
    REP(i, n){
      A[i] = (a[i] >> p) & 1;
      B[i] = (b[i] >> p) & 1;
    }
    vector<int> e;
    e = solve(B, A, n);
    r0[p].resize(n, false);
    for(auto x: e) r0[p][x] = true;
    REP(i, n){
      B[i] = ((b[i] >> p) & 1) ^ 1;
    }
    e = solve(B, A, n);
    r1[p].resize(n, false);
    for(auto x: e) r1[p][x] = true;
  }
  REP(i, n){
    bool ok = true;
    int d = 0;
    REP(p, 30){
      if(r0[p][i]){
        ;
      }else if(r1[p][i]){
        d |= 1 << p;
      }else{
        ok = false;
      }
    }
    if(ok){
      cout << i << " " << d << 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;

#define REP(i,n) for(int i = 0; i < (int)(n); i++)

vector<int> solve(vector<int> a, vector<int> b, int n){
  REP(i, n){
    b.push_back(b[i]);
  }
  a.push_back(2);
  vector<int> e;
  vector<int> t(n + 1);
  t[0] = -1, t[1] = 0;
  int p = 2;
  int q = 0;
  while(p < n + 1){
    if(a[p - 1] == a[q]){
      t[p++] = ++q;
    }else if(q > 0){
      q = t[q];
    }else{
      t[p++] = 0;
    }
  }
  int m = 0;
  p = 0;
  while(m + p < 2 * n){
    if(a[p] == b[m + p]){
      ++p;
    }else{
      m += p - t[p];
      if(p > 0){
        p = t[p];
      }
    }
    if(p == n){
      e.push_back(m);
    }
  }
  return e;
}

int main(){
  int n;
  cin >> n;
  vector<int> a(n), b(n);
  REP(i, n) cin >> a[i];
  REP(i, n) cin >> b[i];
  vector<int> A(n), B(n);
  vector<bool> r0[30];
  vector<bool> r1[30];
  REP(p, 30){
    REP(i, n){
      A[i] = (a[i] >> p) & 1;
      B[i] = (b[i] >> p) & 1;
    }
    vector<int> e;
    e = solve(B, A, n);
    r0[p].resize(n, false);
    for(auto x: e) r0[p][x] = true;
    REP(i, n){
      B[i] = ((b[i] >> p) & 1) ^ 1;
    }
    e = solve(B, A, n);
    r1[p].resize(n, false);
    for(auto x: e) r1[p][x] = true;
  }
  REP(i, n){
    bool ok = true;
    int d = 0;
    REP(p, 30){
      if(r0[p][i]){
        ;
      }else if(r1[p][i]){
        d |= 1 << p;
      }else{
        ok = false;
      }
    }
    if(ok){
      cout << i << " " << d << endl;
    }
  }
  return 0;
}

```