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: Wrong Answer)

#include <iostream>
#include <iomanip>
#include <algorithm>
#include <bitset>
#include <set>
#include <unordered_set>
#include <map>
#include <unordered_map>
#include <cmath>
#include <time.h>
#include <random>
#include <string>
#include <cassert>
#include <vector>
#include <ostream>
#include <istream>
#include <stack>
#include <deque>
#include <queue>
#include <functional>
#include <chrono>
#include <stack>

using namespace std;

#define int long long
#define pb push_back
#define all(a) (a).begin(), (a).end()
#define pii pair<int, int>
#define ld long double

ostream& operator<< (ostream &a, const vector<int> &b) {
    for (auto k : b) cout << k << " ";
    return a;
}

istream& operator>> (istream& a, pii& b) {
    cin >> b.first >> b.second;
    return a;
}

ostream& operator<< (ostream& a, const pii& b) {
    cout << "{" << b.first << ", " << b.second << "}";
    return a;
}

template <typename T1, typename T2> inline void chkmin(T1 &x, const T2 &y) {if (x > y) x = y;}
template <typename T1, typename T2> inline void chkmax(T1 &x, const T2 &y) {if (x < y) x = y;}


#ifdef LOCAL
    #define dbg(x) cout << #x << " : " << (x) << "\n";
    const int INF = 100;
    // const int mod = 2600000069;
    // const int p = 10;
    // const ld PI = 3.1415926535;
#else
    #define dbg(x)
    const int INF = 1e18;
    // const int mod = 2600000069;
    // const int p = 179;
    // const ld PI = 3.1415926535;
#endif

// #pragma GCC optimize("Ofast,no-stack-protector")
// #pragma GCC target("sse,sse2,sse3,sse3,sse4")
// #pragma GCC optimize("unroll-loops")
// #pragma GCC optimize("fast-math")
// #pragma GCC target("avx2")  
// #pragma GCC optimize("section-anchors")
// #pragma GCC optimize("profile-values,profile-reorder-functions,tracer")
// #pragma GCC optimize("vpt")
// #pragma GCC optimize("rename-registers")
// #pragma GCC optimize("move-loop-invariants")
// #pragma GCC optimize("unswitch-loops")
// #pragma GCC optimize("function-sections")
// #pragma GCC optimize("data-sections")

mt19937 gen(chrono::high_resolution_clock::now().time_since_epoch().count());

signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int n;
    cin >> n;
    vector<int> a(n), b(n);
    for (int i = 0; i < n; i++) {
        cin >> a[i];
    }
    for (int i = 0; i < n; i++) {
        cin >> b[i];
    }
    vector<int> ax(n), bx(n), c;
    for (int i = 0; i < n - 1; i++) {
        ax[i] = a[i] ^ a[i + 1];
        bx[i] = b[i] ^ b[i + 1];
    }
    ax[n - 1] = a[n - 1] ^ a[0];
    bx[n - 1] = b[n - 1] ^ b[0];
    c = bx;
    c.pb(INF);
    for (auto k : ax) {
        c.pb(k);
    }
    for (auto k : ax) {
        c.pb(k);
    }
    // dbg(c)
    int sz = c.size();
    vector<int> z(sz, 0);
    int l = 0, r = 0;
    for (int i = 1; i < sz; i++) {
        if (i >= r) {
            while (i + z[i] < sz && c[z[i]] == c[i + z[i]]) {
                z[i]++;
            }
        } else {
            z[i] = min(z[i - l], r - i);
            while (i + z[i] < sz && c[z[i]] == c[i + z[i]]) {
                z[i]++;
            }
        }
        if (i + z[i] >= r) {
            l = i;
            r = i + z[i] + 1;
        }
    }
    vector<int> ans2(n, -1);
    for (int i = 0; i < n; i++) {
        ans2[i] = a[i] ^ b[0];
    }
    for (int i = n + 1; i < n + n + 1; i++) {
        if (z[i] == n) {
            cout << i - n - 1 << " " << ans2[i - n - 1] << "\n";
        }
    }
}
/*

*/