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;
vector<int> t1[2];
vector<int> t2[2];
int a1[100010];
int a2[100010];
int b1[100010];
int b2[100010];
bool t10, t11, t20, t21;
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n - 1; i++) {
    int tmp;
    scanf("%d", &tmp);
    a1[i] = tmp % 2;
    a2[i] = tmp / 2;
  }
  for (int i = 1; i <= n - 1; i++) {
    int tmp;
    scanf("%d", &tmp);
    b1[i] = tmp % 2;
    b2[i] = tmp / 2;
  }
  t10 = t11 = t20 = t21 = true;
  t1[0].push_back(0);
  t1[1].push_back(1);
  for (int i = 1; i <= n - 1; i++) {
    for (int j = 0; j <= 1; j++) {
      if (t10 == true && ((t1[1][i - 1] | j) == a1[i]) &&
          (t1[0][i - 1] & j) == b1[i]) {
        t1[0].push_back(j);
      }
      if (t11 == true && ((t1[1][i - 1] | j) == a1[i]) &&
          ((t1[1][i - 1] & j) == b1[i])) {
        t1[1].push_back(j);
      }
    }
    if (t1[0].size() < i + 1) t10 = false;
    if (t1[1].size() < i + 1) t11 = false;
  }
  t2[0].push_back(0);
  t2[1].push_back(1);
  for (int i = 1; i <= n - 1; i++) {
    for (int j = 0; j <= 1; j++) {
      if (t20 == true && (t2[0][i - 1] | j) == a2[i] &&
          (t2[0][i - 1] & j) == b2[i]) {
        t2[0].push_back(j);
      }
      if (t21 == true && (t2[1][i - 1] | j) == a2[i] &&
          (t2[1][i - 1] & j) == b2[i]) {
        t2[1].push_back(j);
      }
    }
    if (t2[0].size() < i + 1) t20 = false;
    if (t2[1].size() < i + 1) t21 = false;
  }
  if (t1[0].size() < n && t1[1].size() < n ||
      t2[0].size() < n && t2[1].size() < n) {
    printf("NO\n");
    return 0;
  } else {
    printf("YES\n");
    int flag1 = 0, flag2 = 0;
    for (int i = 0; i <= 1; i++) {
      if (t1[i].size() == n) {
        flag1 = i;
        break;
      }
    }
    for (int i = 0; i <= 1; i++) {
      if (t2[i].size() == n) {
        flag2 = i;
        break;
      }
    }
    for (int i = 1; i <= n; i++) {
      printf("%d ", t1[flag1][i - 1] + t2[flag2][i - 1] * 2);
    }
  }
}
