Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

3
0 2 1
1 2 3

Sample Output 1

1 3

If (k,x)=(1,3),

a_0'=(a_1\ XOR \ 3)=1

a_1'=(a_2\ XOR \ 3)=2

a_2'=(a_0\ XOR \ 3)=3

and we have a' = b.

Sample Input 2

5
0 0 0 0 0
2 2 2 2 2

Sample Output 2

0 2
1 2
2 2
3 2
4 2

Sample Input 3

6
0 1 3 7 6 4
1 5 4 6 2 3

Sample Output 3

2 2
5 5

Sample Input 4

2
1 2
0 0

Sample Output 4

No pairs may satisfy the condition.

Solution in C++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include<algorithm>
#include<bitset>
#include<cassert>
#include<cfloat>
#include<climits>
#include<cmath>
#include<deque>
#include<functional>
#include<iomanip>
#include<iostream>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<string>
#include<unordered_map>
#include<unordered_set>
#include<utility>
#include<vector>

using namespace std;

using lint = long long;
using P = pair<int, int>;
using LLP = pair<long long, long long>;

#define REP(i, x, n) for(int i = (x), i##_len = (int)(n) ; i < i##_len ; ++i)
#define rep(i, n) for(int i = 0, i##_len = (int)(n) ; i < i##_len ; ++i)
#define reps(i, n) for(int i = 1, i##_len = (int)(n) ; i <= i##_len ; ++i)
#define rrep(i, n) for(int i = (int)(n) - 1 ; i >= 0 ; --i)
#define rreps(i, n) for(int i = (int)(n) ; i > 0 ; --i)
#define SORT(x) sort((x).begin(), (x).end())
#define SORT_INV(x) sort((x).rbegin(), (x).rend())
#define REVERSE(x) reverse((x).begin(), (x).end())
#define TWINS(x) cout << ((x) ? "Yay!" : ":(") << '\n'
#define endl '\n'

constexpr int IINF = (1 << 30) - 1;
constexpr long long LLINF = 1LL << 61;
constexpr double EPS = 1e-10;

constexpr int dx4[] = {1, 0, -1, 0}, dy4[] = {0, 1, 0, -1};
constexpr int dx8[] = {1, 1, 0, -1, -1, -1, 0, 1}, dy8[] = {0, -1, -1, -1, 0, 1, 1, 1};

template<typename T>
bool chmax(T& a, T b){
    if(a < b){
        a = b;
        return true;
    }
    return false;
}

template<typename T>
bool chmin(T& a, T b){
    if(b < a){
        a = b;
        return true;
    }
    return false;
}

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(10);

    int n;
    cin >> n;

    vector<int> a(n);
    bool ae = true;
    rep(i, n){
        cin >> a[i];
        if(i) ae &= a[i] == a[i - 1];
    }

    vector<int> b(n);
    bool be = true;
    rep(i, n){
        cin >> b[i];
        if(i) be &= b[i] == b[i - 1];
    }

    if(ae && be){
        rep(k, n){
            cout << k << ' ' << (a[0] ^ b[0]) << endl; 
        }
        cout << flush;
        return 0;
    }else if(ae ^ be){
        return 0;
    }

    vector< P > ans;
    rep(k, n){
        int x = a[k] ^ b[0];
        bool flag = true;
        REP(i, 1, n){
            if((a[(i + k) % n] ^ x) != b[i]){
                flag = false;
                break;
            }
        }
        if(flag){
            ans.emplace_back(k, x);
        }
    }

    for(auto p : ans){
        cout << p.first << ' ' << p.second << endl;
    }

    cout << flush;

    return 0;
}