Curiosity Has No Limits

When Masha came to math classes today, she saw two integer sequences of length n−1n−1 on the blackboard. Let’s denote the elements of the first sequence as aiai (0≤ai≤3), and the elements of the second sequence as bibi (0≤bi≤3).

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

• ai=ti|ti+1 (where | denotes the bitwise OR operation) and
• bi=ti&ti+1(where &denotes the bitwise AND operation).

The question appeared to be too difficult for Masha, so now she asked you to check whether such a sequence titi of length nn 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 nn (2≤n≤1052≤n≤105) — the length of the sequence titi.

The second line contains n−1n−1 integers a1,a2,…,an−1a1,a2,…,an−1 (0≤ai≤3) — the first sequence on the blackboard.

The third line contains n−1n−1 integers b1,b2,…,bn−1b1,b2,…,bn−1 (0≤bi≤3) — the second sequence on the blackboard.

Output

In the first line print “YES” (without quotes), if there is a sequence titi 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 nn integers t1,t2,…,tnt1,t2,…,tn (0≤ti≤30≤ti≤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:

• t1|t2=(01)|(11)=(11)=3=a1 and t1&t2=(01)&(11)=(01)=1=b1;
• t2|t3=(11)|(10)=(11)=3=a2 and t2&t3=(11)&(10)=(10)=2=b2;
• t3|t4=(10)|(00)=(10)=2=a3 and t3&t4=(10)&(00)=(00)=0=b3.

In the second example there is no such sequence.

題意：給你a[],b[]，讓你求出t[]. a[i]= t[i] | t[i+1]， b[i] = t[i] & t[i+1].滿足這兩條規則，且t[i]的範圍只能說0，1，2。如果有解，輸出"YES", 並輸出t[i]。

思路：

dfs，先他給第一位的值，然後開始一位一位開始搜尋， 噹噹前位滿足條件時，搜尋下一位，搜尋到最後輸出即可。

程式碼：

``

#include<bits/stdc++.h> 
using namespace std;
int a[200000], b[200000]; 
int t;
int p[200000];
void dfs(int step, int k)
{
	if(step == t) {
		cout << "YES" << endl;
		for(int i = 1; i <= t; i++)
				cout << p[i] << " ";		
			cout << endl;
		exit(0);
	}
		
	for(int i = 0; i <= 3; i++) {
		if(((k | i) == a[step]) && ((k & i) == b[step])) {
			p[step+1] = i;
			dfs(step+1, i);
		}
	}

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

	for(int i = 0; i <= 3; i++) {

		p[1] = i;
		dfs(1, i);
		
	}	 

		cout << "NO" << endl;
	return 0;
}

``