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.
#include <bits/stdc++.h>

using namespace std;

#define int long long
#define f first
#define s second

map <string, int> m;

vector <int> a(100001);
vector <int> b(100001);
vector <int> t(100002);
int n;

void per(int x)
{
	if (x == n)
	{
		cout << "YES" << endl;
		for (int i = 0; i < n; i++)
		{
			cout << t[i] << " ";
		}
		exit(0);
	}
	else
	{
		int aa = a[x - 1];
		int bb = b[x - 1];
		if (!(aa == 3 && bb == 0))
		{
			if (aa == t[x - 1])
			{
				t[x] = bb;
				per(x + 1);
			}
			if (bb == t[x - 1])
			{
				t[x] = aa;
				per(x + 1);
			}
		}
		else
		{
			if (aa == t[x - 1])
			{
				t[x] = bb;
				per(x + 1);
			}
			if (bb == t[x - 1])
			{
				t[x] = aa;
				per(x + 1);
			}

			if (1 == t[x - 1])
			{
				t[x] = 2;
				per(x + 1);
			}
			if (2 == t[x - 1])
			{
				t[x] = 1;
				per(x + 1);
			}
		}
	}
}

main() 
{
	//ios::sync_with_studio(false);
	const int INF = 1e18;
	
	cin >> n;


	m["31"] = 1;
	m["32"] = 1;
	m["30"] = 1;
	m["20"] = 1;
	m["10"] = 1;

	for (int i = 0; i < n - 1; i++)
	{
		cin >> a[i];
	}
	int us = 0;
	for (int i = 0; i < n - 1; i++)
	{
		cin >> b[i];
		string str;
		str = a[i] + '0';
		str += b[i] + '0';
		if (m[str] == 0)
			us = 1;
	}
	if (us == 1)
	{
		cout << "NO";
		return 0;
	}


	if (!(a[0] == 3 && b[0] == 0))
	{
		t[0] = a[0];
		t[1] = b[0];
		per(2);
		t[0] = b[0];
		t[1] = a[0];
		per(2);
	}
	else
	{
		t[0] = a[0];
		t[1] = b[0];
		per(2);
		t[0] = b[0];
		t[1] = a[0];
		per(2);

		t[0] = 2;
		t[1] = 1;
		per(2);
		t[0] = 1;
		t[1] = 2;
		per(2);
	}


	return 0;	
}