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
         
          #define ll long long#define maxn 200005using 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;}