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.
import java.io.BufferedReader;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class nB {

	public void run() throws IOException {
		br = new BufferedReader(new InputStreamReader(System.in));
		pw = new PrintWriter(System.out);
		int n = nextInt() - 1;
		int[] a0 = new int[n];
		int[] a1 = new int[n];
		for (int i = 0; i < n; i++) {
			a0[i] = nextInt();
			a1[i] = a0[i] / 2;
			a0[i] = a0[i] % 2;
		}
		int[] b0 = new int[n];
		int[] b1 = new int[n];
		for (int i = 0; i < n; i++) {
			b0[i] = nextInt();
			b1[i] = b0[i] / 2;
			b0[i] = b0[i] % 2;
		}
		int[] d0 = new int[n];
		for (int i = 0; i < n; i++) {
			int x = a0[i] * 2 + b0[i];
			if (x == 1) {
				pw.println("NO");
				pw.close();
				return;
			}
			if (x == 2) {
				d0[i] = 1;
			}
			if (x == 3) {
				d0[i] = 2;
			}
		}
		int[] res0 = new int[n + 1];
		for (int i = 0; i < n + 1; i++) {
			res0[i] = -1;
		}

		for (int i = 0; i < n; i++) {
			if (d0[i] == 0) {
				if (res0[i] == 1) {
					pw.println("NO");
					pw.close();
					return;
				} else {
					res0[i] = 0;
					res0[i + 1] = 0;
				}
			}
			if (d0[i] == 1) {
				if (res0[i] == 0) {
					res0[i + 1] = 1;
				}
				if (res0[i] == 1) {
					res0[i + 1] = 0;
				}
			}
			if (d0[i] == 2) {
				if (res0[i] == 0) {
					pw.println("NO");
					pw.close();
					return;
				} else {
					res0[i] = 1;
					res0[i + 1] = 1;
				}
			}
		}
		if (res0[n] == -1) {
			res0[n] = 0;
		}
		for (int i = n - 1; i >= 0; i--) {
			if (res0[i] == -1) {
				res0[i] = (res0[i + 1] + 1) % 2;
			}
		}

		int[] d1 = new int[n];
		for (int i = 0; i < n; i++) {
			int x = a1[i] * 2 + b1[i];
			if (x == 1) {
				pw.println("NO");
				pw.close();
				return;
			}
			if (x == 2) {
				d1[i] = 1;
			}
			if (x == 3) {
				d1[i] = 2;
			}
		}
		int[] res1 = new int[n + 1];
		for (int i = 0; i < n + 1; i++) {
			res1[i] = -1;
		}

		for (int i = 0; i < n; i++) {
			if (d1[i] == 0) {
				if (res1[i] == 1) {
					pw.println("NO");
					pw.close();
					return;
				} else {
					res1[i] = 0;
					res1[i + 1] = 0;
				}
			}
			if (d1[i] == 1) {
				if (res1[i] == 0) {
					res1[i + 1] = 1;
				}
				if (res1[i] == 1) {
					res1[i + 1] = 0;
				}
			}
			if (d1[i] == 2) {
				if (res1[i] == 0) {
					pw.println("NO");
					pw.close();
					return;
				} else {
					res1[i] = 1;
					res1[i + 1] = 1;
				}
			}
		}
		if (res1[n] == -1) {
			res1[n] = 0;
		}
		for (int i = n - 1; i >= 0; i--) {
			if (res1[i] == -1) {
				res1[i] = (res1[i + 1] + 1) % 2;
			}
		}
		
		pw.println("YES");
		for (int i = 0; i < n + 1; i++) {
			pw.print((res1[i]*2+res0[i]) + " "); 
		}
		pw.close();
	}

	BufferedReader br;
	PrintWriter pw;
	StringTokenizer st;

	public String nextToken() throws IOException {
		while (st == null || !st.hasMoreTokens()) {
			st = new StringTokenizer(br.readLine());
		}
		return st.nextToken();
	}

	public int nextInt() throws IOException {
		return Integer.parseInt(nextToken());
	}

	public double nextDouble() throws IOException {
		return Double.parseDouble(nextToken());
	}

	public static void main(String[] args) throws IOException {
		new nB().run();
	}

}