When Masha came to math classes today, she saw two integer sequences of length n - 1 on the blackboard. Let's denote the elements of the first sequence as a_i (0 ≤ a_i ≤ 3), and the elements of the second sequence as b_i (0 ≤ b_i ≤ 3).

Masha became interested if or not there is an integer sequence of length n, which elements we will denote as t_i (0 ≤ t_i ≤ 3), so that for every i (1 ≤ i ≤ n - 1) the following is true: 

  * a_i = t_i | t_{i + 1} (where | denotes the [bitwise OR operation](https://en.wikipedia.org/wiki/Bitwise_operation#OR)) and 
  * b_i = t_i \& t_{i + 1} (where \& denotes the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND)). 



The question appeared to be too difficult for Masha, so now she asked you to check whether such a sequence t_i of length n exists. If it exists, find such a sequence. If there are multiple such sequences, find any of them.

Input

The first line contains a single integer n (2 ≤ n ≤ 10^5) — the length of the sequence t_i. 

The second line contains n - 1 integers a_1, a_2, …, a_{n-1} (0 ≤ a_i ≤ 3) — the first sequence on the blackboard.

The third line contains n - 1 integers b_1, b_2, …, b_{n-1} (0 ≤ b_i ≤ 3) — the second sequence on the blackboard.

Output

In the first line print "YES" (without quotes), if there is a sequence t_i that satisfies the conditions from the statements, and "NO" (without quotes), if there is no such sequence.

If there is such a sequence, on the second line print n integers t_1, t_2, …, t_n (0 ≤ t_i ≤ 3) — the sequence that satisfies the statements conditions.

If there are multiple answers, print any of them.

Examples

Input

4
3 3 2
1 2 0


Output

YES
1 3 2 0 

Input

3
1 3
3 2


Output

NO

Note

In the first example it's easy to see that the sequence from output satisfies the given conditions: 

  * t_1 | t_2 = (01_2) | (11_2) = (11_2) = 3 = a_1 and t_1 \& t_2 = (01_2) \& (11_2) = (01_2) = 1 = b_1; 
  * t_2 | t_3 = (11_2) | (10_2) = (11_2) = 3 = a_2 and t_2 \& t_3 = (11_2) \& (10_2) = (10_2) = 2 = b_2; 
  * t_3 | t_4 = (10_2) | (00_2) = (10_2) = 2 = a_3 and t_3 \& t_4 = (10_2) \& (00_2) = (00_2) = 0 = b_3. 



In the second example there is no such sequence.
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse2")
 
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
#define ll                long long
#define float             long double
#define pb                push_back
#define ppb               pop_back
#define lb                lower_bound
#define ub                upper_bound
#define mii               map<int,int>
#define pii               pair<int,int>
#define vi                vector<int>
#define all(x)            (x).begin(),(x).end()
#define mod               1000000007
#define hell              998244353
#define fr                first
#define sc                second
#define rep(i,a,b)        for(int i=a;i<b;i++)
#define mem1(a)           memset(a,-1,sizeof(a))
#define mem0(a)           memset(a,0,sizeof(a))
#define chrng(n,a,b)      (((n)>=(a))&&((n)<=(b)))
#define clamp(n,a,b)      (((n)<(a))?(a):((((n)>(b))?(b):(n)))
map<char,int> mp;
bool sortbysec(const pair<int,int> &a,const pair<int,int> &b) 
{ 
    return (a.second > b.second); 
}


int solve()
{
	int n,i,j;
	cin>>n;
	vector<int>vec(n);
	int m=n-1;
	int arr[m],brr[m];
	for(i=0;i<m;i++)cin>>arr[i];
	for(i=0;i<m;i++)cin>>brr[i];
	
	int ca=0;
	for(i=0;i<=3;i++)
	{
		for(j=0;j<=3;j++)
		{
			if(((i|j)==arr[0])&&((i&j)==brr[0]))
			{
				ca=1;
				break;
			}
		}
		if(ca)break;
	}
	if(!ca)
	{cout<<"NO";return 0;}
	vec[0]=j,vec[1]=i;
	if(n>2)
	{
		for(i=0;i<=3;i++)
    	{
	    	if(((vec[0]|i)==arr[1])&&((vec[0]&i)==brr[1]))
	     	{swap(vec[0],vec[1]);break;}
    	}
	}
	for(i=1;i<m;i++)
	{
		ca=0;
		for(j=0;j<=3;j++)
		{
			if(((vec[i]|j)==arr[i])&&((vec[i]&j)==brr[i]))
        	{
		       	vec[i+1]=j;
        		ca=1;
		       	break;
        	}
        }
		if(!ca)
		{
			cout<<"NO";
			return 0;
		}
	}
	cout<<"YES\n";
	for(i=0;i<n;i++)
	cout<<vec[i]<<" ";
}
signed main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	//freopen("input.txt", "r", stdin);
	//freopen("output.txt", "w", stdout);
	#ifdef SEIVE
		seive();
	#endif
	#ifdef NCR
		init();
	#endif
	
	int t=1;
//	cin>>t;
	while(t--) 
	solve();
	return 0;
}