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;
long long int *arr;
bool subs_exists(int start, int end);
int main() {
  int n;
  cin >> n;
  int a[n - 1], b[n - 1];
  for (int i = 0; i < n - 1; i++) {
    cin >> a[i];
  }
  for (int i = 0; i < n - 1; i++) {
    cin >> b[i];
  }
  int c0[n], c1[n];
  c0[0] = 1;
  int flag = 0;
  for (int i = 0; i < n - 1; i++) {
    if (c0[i] == 0) {
      if (a[i] % 2 == 0) {
        if (b[i] % 2 == 0)
          c0[i + 1] = 0;
        else {
          flag = 1;
          break;
        }
      } else {
        if (b[i] % 2 == 0)
          c0[i + 1] = 1;
        else
          c0[i + 1] = 1;
      }
    } else {
      if (a[i] % 2 == 0) {
        flag = 1;
        break;
      } else {
        if (b[i] % 2 == 0)
          c0[i + 1] = 0;
        else
          c0[i + 1] = 1;
      }
    }
  }
  if (flag == 1) {
    flag = 0;
    c0[0] = 0;
    for (int i = 0; i < n - 1; i++) {
      if (c0[i] == 0) {
        if (a[i] % 2 == 0) {
          if (b[i] % 2 == 0)
            c0[i + 1] = 0;
          else {
            flag = 1;
            break;
          }
        } else {
          if (b[i] % 2 == 0)
            c0[i + 1] = 1;
          else
            c0[i + 1] = 1;
        }
      } else {
        if (a[i] % 2 == 0) {
          flag = 1;
          break;
        } else {
          if (b[i] % 2 == 0)
            c0[i + 1] = 0;
          else
            c0[i + 1] = 1;
        }
      }
    }
    if (flag == 1) {
      cout << "NO" << endl;
      return 0;
    }
  }
  c1[0] = 1;
  for (int i = 0; i < n - 1; i++) {
    if (c1[i] == 0) {
      if ((a[i] / 2) % 2 == 0) {
        if ((b[i] / 2) % 2 == 0)
          c1[i + 1] = 0;
        else {
          flag = 1;
          break;
        }
      } else {
        if ((b[i] / 2) % 2 == 0)
          c1[i + 1] = 1;
        else
          c1[i + 1] = 1;
      }
    } else {
      if ((a[i] / 2) % 2 == 0) {
        flag = 1;
        break;
      } else {
        if ((b[i] / 2) % 2 == 0)
          c1[i + 1] = 0;
        else
          c1[i + 1] = 1;
      }
    }
  }
  if (flag == 1) {
    flag = 0;
    c1[0] = 0;
    for (int i = 0; i < n - 1; i++) {
      if (c1[i] == 0) {
        if ((a[i] / 2) % 2 == 0) {
          if ((b[i] / 2) % 2 == 0)
            c1[i + 1] = 0;
          else {
            flag = 1;
            break;
          }
        } else {
          if ((b[i] / 2) % 2 == 0)
            c1[i + 1] = 1;
          else
            c1[i + 1] = 1;
        }
      } else {
        if ((a[i] / 2) % 2 == 0) {
          flag = 1;
          break;
        } else {
          if ((b[i] / 2) % 2 == 0)
            c1[i + 1] = 0;
          else
            c1[i + 1] = 1;
        }
      }
    }
    if (flag == 1) {
      cout << "NO" << endl;
      return 0;
    }
  }
  cout << "YES" << endl;
  for (int i = 0; i < n; i++) cout << c0[i] + 2 * c1[i] << " ";
  cout << endl;
}
bool subs_exists(int start, int end) {
  int sum = arr[end + 1] - arr[start];
  int n = end + 1 - start;
  if (sum % 2 != 0) return false;
  bool **dp = (bool **)malloc(sizeof(bool *) * (sum / 2 + 1));
  for (int i = 0; i < sum / 2 + 1; i++) {
    dp[i] = (bool *)malloc(sizeof(bool *) * n);
  }
  for (int i = 0; i < n; i++) {
    dp[0][i] = true;
  }
  for (int i = 1; i < sum / 2 + 1; i++) {
    for (int j = 0; j < n; j++) {
      if (j != 0 && dp[i][j - 1]) {
        dp[i][j] = true;
      }
      if (j != 0 && i - (arr[j + start + 1] - arr[start]) >= 0 &&
          dp[i - (arr[j + start + 1] - arr[start])][j - 1]) {
        dp[i][j] = true;
      }
      if (arr[j + start + 1] - arr[start] == i) {
        dp[i][j] = true;
      }
    }
  }
  return dp[sum / 2][n - 1];
}
