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;
struct mat {
  int a;
  int o;
} match[5][5];
int n, a[100100], b[100100];
int test1[100100], test2[100100];
int main() {
  memset(match, -1, sizeof(match));
  for (int i = 0; i <= 3; ++i) {
    for (int j = i; j <= 3; ++j) {
      int x, y;
      x = i | j;
      y = i & j;
      match[x][y].o = i, match[x][y].a = j;
    }
  }
  while (cin >> n) {
    memset(a, 0, sizeof(a));
    memset(b, 0, sizeof(b));
    for (int i = 1; i < n; ++i) cin >> a[i];
    for (int i = 1; i < n; ++i) cin >> b[i];
    int flag = 1;
    int t1[100100], t2[100100];
    memset(t1, -1, sizeof(t1));
    memset(t2, -1, sizeof(t2));
    for (int i = 1; i < n; ++i) {
      t1[i] = match[a[i]][b[i]].o;
      t2[i] = match[a[i]][b[i]].a;
      if (t1[i] == -1 || t2[i] == -1) {
        flag = 0;
        break;
      }
    }
    if (n == 1) {
      cout << t1[1] << ' ' << t2[1] << endl;
      continue;
    }
    int ok1, ok2;
    ok1 = ok2 = 1;
    memset(test1, 0, sizeof(test1));
    memset(test2, 0, sizeof(test2));
    test1[1] = t1[1], test1[2] = t2[1];
    test2[1] = t2[1], test2[2] = t1[1];
    for (int i = 2; i < n; ++i) {
      if (t1[i] == test1[i]) {
        test1[i] = t1[i];
        test1[i + 1] = t2[i];
      } else if (t2[i] == test1[i]) {
        test1[i] = t2[i];
        test1[i + 1] = t1[i];
      } else {
        ok1 = 0;
        break;
      }
    }
    for (int i = 2; i < n; ++i) {
      if (t1[i] == test2[i]) {
        test2[i] = t1[i];
        test2[i + 1] = t2[i];
      } else if (t2[i] == test2[i]) {
        test2[i] = t2[i];
        test2[i + 1] = t1[i];
      } else {
        ok2 = 0;
        break;
      }
    }
    if (ok1 == 0 && ok2 == 0) flag = 0;
    if (flag == 0)
      cout << "NO\n";
    else {
      cout << "YES\n";
      if (ok1) {
        for (int i = 1; i <= n; ++i) cout << test1[i] << ' ';
        getchar();
        cout << endl;
        continue;
      } else {
        for (int i = 1; i <= n; ++i) cout << test2[i] << ' ';
        getchar();
        cout << endl;
      }
    }
  }
  return 0;
}
