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.DataInputStream;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.ArrayList;

public class B {
    static class Parser {
        final private int BUFFER_SIZE = 1 << 16;
        private DataInputStream din;
        private byte[] buffer;
        private int bufferPointer, bytesRead;

        public Parser(InputStream in) {
            din = new DataInputStream(in);
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }

        public String nextString(int maxSize) {
            byte[] ch = new byte[maxSize];
            int point = 0;
            try {
                byte c = read();
                while (c == ' ' || c == '\n' || c == '\r')
                    c = read();
                while (c != ' ' && c != '\n' && c != '\r') {
                    ch[point++] = c;
                    c = read();
                }
            } catch (Exception e) {
            }
            return new String(ch, 0, point);
        }

        public int nextInt() {
            int ret = 0;
            boolean neg;
            try {
                byte c = read();
                while (c <= ' ')
                    c = read();
                neg = c == '-';
                if (neg)
                    c = read();
                do {
                    ret = ret * 10 + c - '0';
                    c = read();
                } while (c > ' ');

                if (neg) return -ret;
            } catch (Exception e) {
            }
            return ret;
        }

        public long nextLong() {
            long ret = 0;
            boolean neg;
            try {
                byte c = read();
                while (c <= ' ')
                    c = read();
                neg = c == '-';
                if (neg)
                    c = read();
                do {
                    ret = ret * 10 + c - '0';
                    c = read();
                } while (c > ' ');

                if (neg) return -ret;
            } catch (Exception e) {
            }
            return ret;
        }

        private void fillBuffer() {
            try {
                bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
            } catch (Exception e) {
            }
            if (bytesRead == -1) buffer[0] = -1;
        }

        private byte read() {
            if (bufferPointer == bytesRead) fillBuffer();
            return buffer[bufferPointer++];
        }
    }

    public static void main(String[] args) {
        Parser in = new Parser(System.in);
        PrintWriter out = new PrintWriter(System.out);
        long c = 0;
        int n = in.nextInt();
        int a[] = new int[n - 1];
        int b[] = new int[n - 1];
        for (int i = 0; i < n - 1; i++) {
            a[i] = in.nextInt();
        }
        for (int i = 0; i < n - 1; i++) {
            b[i] = in.nextInt();
        }
        ArrayList<Integer> start = new ArrayList<>();
        for (int i = 0; i < 4; i++) {
            for (int j = 0; j < 4; j++) {
                if ((i & j) == b[0] && (i | j) == a[0]) start.add(i * 100 + j);
            }
        }
        if (start.size() == 0) {
            System.out.println("NO");
            return;
        }
        int t[] = new int[n];
        for (int i = 0; i < start.size(); i++) {
            t[0] = start.get(i) / 100;
            t[1] = start.get(i) % 100;
            boolean succ = true;
            for (int j = 2; j < n; j++) {
                boolean br = false;
                for (int k = 0; k < 4; k++) {
                    if((k & t[j-1]) == b[j-1] && (k | t[j-1]) == a[j-1]){
                        t[j] = k;
                        br = true;
                    }
                }
                if(!br){
                    succ = false;
                    break;
                }
            }
            if(succ){
                out.println("YES");
                for (int j = 0; j < n; j++) {
                    out.print(t[j]+" ");
                }
                out.close();
                return;
            }
        }
        System.out.println("NO");
    }
}
