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)

#pragma GCC optimize("O3")
#include<bits/stdc++.h> 
using namespace std;
using ll=long long;
typedef  unsigned long long int ull;
typedef pair<ll,ll> P;
template<class T> using V=vector<T>; 
#define fi first
#define se second
#define all(v) (v).begin(),(v).end()
const ll inf=(1e18);
//const ll mod=998244353;
const ll mod=1000000007;
//ios_base::sync_with_stdio(false);
//cin.tie(NULL);
ll gcd(ll a,ll b) {return b ? gcd(b,a%b):a;}
ll lcm(ll c,ll d){return c/gcd(c,d)*d;}
struct __INIT{__INIT(){cin.tie(0);ios::sync_with_stdio(false);cout<<fixed<<setprecision(15);}} __init;
template<class T> bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T> bool chmin(T &a, const T &b) { if (a>b) { a=b; return 1; } return 0; }
V<int> z_algo(const V<int> &s){
    int n=s.size();
    V<int> prefix(n);
    for(int i=1,j=0;i<n;i++){
        if(i+prefix[i-j]<j+prefix[j]){
            //すでに計算済みの場所である
            prefix[i]=prefix[i-j];
        }else{
            // わからないところを愚直に調べる
            int k=max(0,j+prefix[j]-i);
            while(i+k<n&&s[k]==s[i+k])k++;
            prefix[i]=k;
            j=i;
        }
    }
    prefix[0]=n;
    return prefix;
}
int main(){
    int n;
	cin>>n;
	V<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];
	V<int> s(3*n);
	for(int i=0;i<n;i++){
		s[i]=(b[i]^b[(i+1)%n]);
	}
	for(int i=0;i<n;i++){
		s[i+n]=(a[i]^a[(i+1)%n]);
		s[i+2*n]=(a[i]^a[(i+1)%n]);
	}
	V<int> res=z_algo(s);
	int cnt=0;
	for(int i=n;i<2*n;i++){
		if(res[i]>=n){
			cout<<i-n<<" "<<(a[i-n]^b[0])<<endl;
		}
	}
}