Anton and School

Anton and School

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 bi (0 ≤ bi ≤ 109) — elements of the array b.

Third line contains n integers ci (0 ≤ ci ≤ 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 ai — elements of the array a. If there are multiple possible solutions, you may print any of them.

Examples

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
分析：可以证明答案数组是a[i]=（b[i]+c[i]+Σ(b[i]+c[i]）/（2n））/n;
　　　然后根据异或合取性质检验答案；
代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <unordered_map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
#define pii pair<int,int>
#define Lson L, mid, ls[rt]
#define Rson mid+1, R, rs[rt]
#define sys system("pause")
#define freopen freopen("in.txt","r",stdin)
const int maxn=2e5+10;
using namespace std;
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
inline ll read()
{
    ll x=0;int f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n,m,t;
ll a[maxn],b[maxn],c[maxn],d[maxn],k[maxn],num[30][maxn],sum;
bool flag;
int main()
{
    int i,j;
    flag=true;
    scanf("%d",&n);
    rep(i,1,n)b[i]=read();
    rep(i,1,n)c[i]=read();
    rep(i,1,n)d[i]=b[i]+c[i],sum+=d[i];
    sum/=(2*n);
    rep(i,1,n)a[i]=(d[i]-sum)/n;
    rep(i,1,n)
    {
        rep(j,0,29)if((a[i]>>j)&1)num[j][i]=1,k[j]++;
    }
    rep(i,1,n)
    {
        ll tmp1=0,tmp2=0;
        rep(j,0,29)
        {
            if(num[j][i])tmp1+=(1<<j)*k[j];
            else tmp2+=(1<<j)*k[j];
        }
        if(tmp1!=b[i]&&tmp2!=c[i])
        {
            flag=false;
            break;
        }
    }
    if(flag)
    {
        rep(i,1,n)printf("%lld ",a[i]);
    }
    else puts("-1");
    //system("Pause");
    return 0;
}