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
#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for(int i = 0; i < n; i++)
#define rep2(i, x, n) for(int i = x; i <= n; i++)
#define rep3(i, x, n) for(int i = x; i >= n; i--)
#define elif else if
typedef long long ll;
typedef pair<ll, ll> P;
const ll MOD = 1e9+7;
const ll MOD2 = 998244353;
const ll INF = 1e18;
const string alpha = "abcdefghijklmnopqrstuvwxyz";

int main(){
    int N;
    cin >> N;
    ll a[N], b[N];
    rep(i, N) cin >> a[i];
    rep(i, N) cin >> b[i];
    ll A[2*N], B[N];
    rep(i, N-1){
        A[i] = a[i]^a[i+1];
        A[N+i] = A[i];
        B[i] = b[i]^b[i+1];
    }
    A[N-1] = a[N-1]^a[0];
    B[N-1] = b[N-1]^b[0];
    ll mix[3*N-1];
    rep(i, N) mix[i] = B[i];
    rep2(i, N, 3*N-2) mix[i] = A[i-N];
    ll cnt[3*N-1];
    cnt[0] = 3*N-1;
    int i = 1, j = 0;
    while(i < 3*N-1){
        while(i+j < 3*N-1 && mix[j] == mix[i+j]) j++;
        cnt[i] = j;
        if(j == 0) {i++; continue;}
        int k = 1;
        while(i+k < 3*N-1 && k+cnt[k] < j) cnt[i+k] = cnt[k], k++;
        i += k, j -= k;
    }
    rep2(i, N, 2*N-1){
        if(cnt[i] >= N-1){
            ll ans = a[i%N]^b[0];
            cout << i%N << ' ' << ans << endl;
        }
    }
}