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>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

#define all(a) a.begin(),a.end()
#define INT_INF 0x3f3f3f3f
#define LL_INF 0x3f3f3f3f3f3f3f3f
#define ofk order_of_key
#define fbo find_by_order
#define mp make_pair
#define pb push_back
#define endl "\n"
#define se second
#define f first

using namespace std;
using namespace __gnu_pbds;

typedef tree<int , null_type , less_equal<int> , rb_tree_tag , tree_order_statistics_node_update> ordered_set;
typedef unsigned long long ull;
typedef long long ll;
typedef long double ld;
 
const int MAXN = (int)(1e5 + 123);

void boost() {
	ios_base::sync_with_stdio(false);
	
	cin.tie(0);

	cout.tie(0);
}
void file() {
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
}
ll gcd(ll a, ll b){
	return b == 0 ? a : gcd(b, a % b);
}
ll lcm(ll a, ll b){
	return a*b/gcd(a,b);
}
ll fpow(ll  b, ll exp, ll mod = LL_INF) {
	if(exp == 0) return 1;
	ll t = fpow(b,exp/2,mod);
	if(exp&1) return t*t%mod*b%mod;
	return t*t%mod;
}
// <+=============================================================================================================+>
int a[MAXN], b[MAXN] , t[MAXN];


int main() {
	file();
	boost();
	
	int n;
	cin >> n;
	for (int i=1; i<n; i++) {
		cin >> a[i];
	}
	for (int i=1; i<n; i++) {
		cin >> b[i];
	}

	for (int st=0; st<=3; st++) {
		t[1] = st;
		int p = 1;
		bool no = false;
		for (int i=1; i<n; i++) {
			bool ok = false;
			for (int v=0; v<=3; v++) {
				if (a[i] == (t[p]|v) && b[i] == (t[p]&v)) {
					ok = true;
					t[++p] = v;
					break;
				}
			}
			if (!ok) {
				no = true;
				break;
			}
		}
		if (!no) {
			cout <<"YES"<<endl;
			for (int i=1; i<=n; i++) {
				cout <<t[i]<<" ";
			}
			return 0;
		}
	}
	cout <<"NO"<<endl;
	return 0;
}
