#include<bits/stdc++.h>
#define pb push_back
#define mp make_pair
#define fst first
#define snd second
using namespace std;

typedef long long ll;
typedef double db;
typedef long double ldb;

template<class T> inline void checkmin(T &a,const T &b){if(b<a) a=b;}
template<class T> inline void checkmax(T &a,const T &b){if(b>a) a=b;}

const int inf=1e9+10;
const int maxn=2e5+10;

int a[maxn],b[maxn],difa[maxn],difb[maxn];
int n;
vector <int> sa,sb,s;

vector <int> build_zfunction(vector <int> &s){
	vector <int> z(s.size(),0);
	z[0]=(int)s.size();
	int j=1;
	for(int i=1;i<(int)s.size();i++){
		int u=max(0,min(j+z[j]-i,z[i-j]));
		while(s[u]==s[i+u]) u++;
		z[i]=u;
		if(i+z[i]>=j+z[j]) j=i;
	}
	return z;
}

int main(){
	scanf("%d",&n);
	for(int i=0;i<n;i++) scanf("%d",&a[i]);
	for(int i=0;i<n;i++) scanf("%d",&b[i]);
	for(int i=0;i<n;i++) difa[i]=a[(i+1)%n]^a[i];
	for(int i=0;i<n;i++) difb[i]=b[(i+1)%n]^b[i];
	for(int i=0;i<n;i++){
		sa.pb(difa[i]);
		sb.pb(difb[i]);
	}
	for(int i=0;i<sb.size();i++) s.pb(sb[i]);
	s.pb(inf);
	for(int i=0;i<sa.size();i++) s.pb(sa[i]);
	for(int i=0;i<sa.size();i++) s.pb(sa[i]);
	vector<int> z=build_zfunction(s),ans;
	int k=0;
	for(int i=(int)sb.size()+1;i<z.size();i++,k++){
		if(k>=n) break;
		if(z[i]==n){
			ans.pb(k);
		}
	}
	for(int i=0;i<ans.size();i++){
		int k=ans[i],x=a[i]^b[(i-k+n)%n];
		printf("%d %d\n",k,x);
	}
	return 0;
}

### Prompt

Your challenge is to write 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>
#define pb push_back
#define mp make_pair
#define fst first
#define snd second
using namespace std;

typedef long long ll;
typedef double db;
typedef long double ldb;

template<class T> inline void checkmin(T &a,const T &b){if(b<a) a=b;}
template<class T> inline void checkmax(T &a,const T &b){if(b>a) a=b;}

const int inf=1e9+10;
const int maxn=2e5+10;

int a[maxn],b[maxn],difa[maxn],difb[maxn];
int n;
vector <int> sa,sb,s;

vector <int> build_zfunction(vector <int> &s){
	vector <int> z(s.size(),0);
	z[0]=(int)s.size();
	int j=1;
	for(int i=1;i<(int)s.size();i++){
		int u=max(0,min(j+z[j]-i,z[i-j]));
		while(s[u]==s[i+u]) u++;
		z[i]=u;
		if(i+z[i]>=j+z[j]) j=i;
	}
	return z;
}

int main(){
	scanf("%d",&n);
	for(int i=0;i<n;i++) scanf("%d",&a[i]);
	for(int i=0;i<n;i++) scanf("%d",&b[i]);
	for(int i=0;i<n;i++) difa[i]=a[(i+1)%n]^a[i];
	for(int i=0;i<n;i++) difb[i]=b[(i+1)%n]^b[i];
	for(int i=0;i<n;i++){
		sa.pb(difa[i]);
		sb.pb(difb[i]);
	}
	for(int i=0;i<sb.size();i++) s.pb(sb[i]);
	s.pb(inf);
	for(int i=0;i<sa.size();i++) s.pb(sa[i]);
	for(int i=0;i<sa.size();i++) s.pb(sa[i]);
	vector<int> z=build_zfunction(s),ans;
	int k=0;
	for(int i=(int)sb.size()+1;i<z.size();i++,k++){
		if(k>=n) break;
		if(z[i]==n){
			ans.pb(k);
		}
	}
	for(int i=0;i<ans.size();i++){
		int k=ans[i],x=a[i]^b[(i-k+n)%n];
		printf("%d %d\n",k,x);
	}
	return 0;
}

```