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: Accepted)

#include <limits.h>
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#include <algorithm>
#include <cassert>
#include <cfloat>
#include <complex>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <regex>
#include <set>
#include <stack>
#include <string>
#include <unordered_map>
#include <vector>

using namespace std;

#define chmax(x, y) x = max(x, y)
#define chmin(x, y) x = min(x, y)
#define rep(i, n) for (ll i = 0; i < (n); ++i)
#define repLRE(i, l, r) for (ll i = (l); i <= (r); ++i)
#define rrepLRE(i, l, r) for (ll i = (l); i >= (r); --i)
#define Sort(v) sort(v.begin(), v.end())
#define rSort(v) sort(v.rbegin(), v.rend())
#define Reverse(v) reverse(v.begin(), v.end())
#define Lower_bound(v, x) \
  distance(v.begin(), lower_bound(v.begin(), v.end(), x))
#define Upper_bound(v, x) \
  distance(v.begin(), upper_bound(v.begin(), v.end(), x))

using ll = long long;
using ull = unsigned long long;
using P = pair<ll, ll>;
using T = tuple<ll, ll, ll>;
using vll = vector<ll>;
using vP = vector<P>;
using vT = vector<T>;
using vvll = vector<vector<ll>>;
using vvP = vector<vector<P>>;
using dqll = deque<ll>;

ll dx[9] = {-1, 1, 0, 0, -1, -1, 1, 1, 0};
ll dy[9] = {0, 0, -1, 1, -1, 1, -1, 1, 0};

/* Macros reg. ends here */

const ll INF = 1LL << 50;

struct RollingHash {
  static const ll base1 = 1007, base2 = 2009;
  static const ll mod1 = 1000000007, mod2 = 1000000009;
  vector<ll> hash1, hash2, power1, power2;

  // construct
  RollingHash(const vll& S) {
    ll n = (ll)S.size();
    hash1.assign(n + 1, 0);
    hash2.assign(n + 1, 0);
    power1.assign(n + 1, 1);
    power2.assign(n + 1, 1);
    for (ll i = 0; i < n; ++i) {
      hash1[i + 1] = (hash1[i] * base1 + S[i]) % mod1;
      hash2[i + 1] = (hash2[i] * base2 + S[i]) % mod2;
      power1[i + 1] = (power1[i] * base1) % mod1;
      power2[i + 1] = (power2[i] * base2) % mod2;
    }
  }

  // get hash of s[l..r-1]
  inline P get(ll l, ll r) const {
    long long res1 = hash1[r] - hash1[l] * power1[r - l] % mod1;
    if (res1 < 0) res1 += mod1;
    long long res2 = hash2[r] - hash2[l] * power2[r - l] % mod2;
    if (res2 < 0) res2 += mod2;
    return {res1, res2};
  }

  // get lcp of s[a..] and s[b..]
  inline int getLCP(int a, int b) const {
    int len = min((int)hash1.size() - a, (int)hash1.size() - b);
    int low = 0, high = len;
    while (high - low > 1) {
      int mid = (low + high) >> 1;
      if (get(a, a + mid) != get(b, b + mid))
        high = mid;
      else
        low = mid;
    }
    return low;
  }
};

int main() {
  // ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cout << fixed << setprecision(15);

  ll n;
  cin >> n;
  vll as(n), bs(n), kcats(3 *n);

  rep(i, n) cin >> as[i];
  rep(i, n) cin >> bs[i];

  rep(i, n) {
    kcats[i] = bs[(i + 1)%n] ^ bs[i];
    kcats[n + i] = kcats[2 * n + i] = as[(i + 1)%n] ^ as[i];
  }

  RollingHash rh(kcats);

  vP ans;
  P hval = rh.get(0, n);
  rep(k, n) {
    P curhval = rh.get(n + k, 2 * n + k);
    if (hval == curhval) {
      ll x = as[k] ^ bs[0];
      ans.emplace_back(k, x);
    }
  }

  Sort(ans);
  for (P p : ans) {
    cout << p.first << " " << p.second << endl;
  }

  return 0;
}