Codeforces 734 F Anton and School

Discription

Anton goes to school, his favorite lessons are arraystudying. He usually solves all the tasks pretty fast, but this time the teacher gave him a complicated one: given two arrays b and c of length n, find array a, such that:

where a and b means bitwise AND, while a or b means bitwise OR.

Usually Anton is good in arraystudying, but this problem is too hard, so Anton asks you to help.

Input

The first line of the input contains a single integers n (1 ≤ n ≤ 200 000) — the size of arrays b and c.

The second line contains n integers b_i (0 ≤ b_i ≤ 10⁹) — elements of the array b.

Third line contains n integers c_i (0 ≤ c_i ≤ 10⁹) — elements of the array c.

Output

If there is no solution, print  - 1.

Otherwise, the only line of the output should contain n non-negative integers a_i — elements of the array a. If there are multiple possible solutions, you may print any of them.

Example

Input

4
6 8 4 4
16 22 10 10

Output

3 5 1 1 

Input

5
8 25 14 7 16
19 6 9 4 25

Output

-1


有一个显然的性质就是 (a&b)+(a|b)=a+b。
这个考虑每一位的贡献就行了。
于是我们可以得到c[i]+b[i]=a[i]*n+∑a[],从而可以开开心心的算出每一个a，然后带回去验证一下就可以了

/*
   We know that b[i]+c[i]=n*a[i]+∑a[] 
*/
#include<bits/stdc++.h>
#define ll long long
#define maxn 200005
using namespace std;
int b[maxn],c[maxn];
int a[maxn],n,m,ci[60];
ll tot=0,now;
int main(){
	ci[0]=1;
	for(int i=1;i<=30;i++) ci[i]=ci[i-1]<<1;
	
	scanf("%d",&n);
	for(int i=1;i<=n;i++) scanf("%d",b+i),tot+=(ll)b[i];
	for(int i=1;i<=n;i++) scanf("%d",c+i),tot+=(ll)c[i];
	//tot is used to calculate the sum of array a[]
	if(tot%(n<<1)){
		puts("-1");
		return 0;
	}
	
	tot/=(n<<1);
	
	for(int i=1;i<=n;i++){
		now=c[i]+b[i]-tot;
		if(now%n){
			puts("-1");
			return 0;
		}
		a[i]=now/n;
	}
	
	for(int i=0;i<=30;i++){
		int cnt=0;
		for(int j=1;j<=n;j++) if(a[j]&ci[i]) cnt++;
		for(int j=1;j<=n;j++){
			if(a[j]&ci[i]) b[j]-=cnt*ci[i],c[j]-=n*ci[i];
			else c[j]-=cnt*ci[i];
		}
	}
	
	for(int i=1;i<=n;i++) if(c[i]||b[i]){
		puts("-1");
		return 0;
	}
	
	for(int i=1;i<=n;i++) printf("%d ",a[i]);
	return 0;
}