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;
int a[100010], b[100010], t[100010];
int tmp[100010];
int n;
bool check() {
  for (int i = 3; i < n; i++) {
    int t1, t2;
    if (a[i] >= b[i])
      t1 = a[i], t2 = b[i];
    else
      return 0;
    if (t1 != tmp[i]) {
      swap(t1, t2);
      if (t1 != tmp[i]) {
        if (t1 == 0 && t2 == 3) {
          t1 = 2, t2 = 1;
          if (t1 != tmp[i]) {
            swap(t1, t2);
            if (t1 != tmp[i]) return 0;
          }
        } else
          return 0;
      }
    }
    tmp[i + 1] = t2;
  }
  return 1;
}
int main() {
  cin >> n;
  for (int i = 1; i < n; i++) scanf("%d", &a[i]);
  for (int i = 1; i < n; i++) scanf("%d", &b[i]);
  for (int i = 1; i < n; i++)
    if (a[i] == 2 && b[i] == 1) a[i] = 3, b[i] = 0;
  if (a[1] < b[1]) {
    printf("NO");
    return 0;
  }
  t[1] = a[1], t[2] = b[1];
  int t1, t2;
  if (a[2] >= b[2])
    t1 = a[2], t2 = b[2];
  else {
    printf("NO");
    return 0;
  }
  int vis = 0;
  if (t1 == 3 && t2 == 0) {
    if (t1 == t[2]) {
      vis = 1;
      t[3] = t2;
      memcpy(tmp, t, sizeof(tmp));
      if (!check()) vis = 0;
    }
    if (!vis) {
      swap(t[2], t[1]);
      if (t1 == t[2]) {
        vis = 1;
        t[3] = t2;
        memcpy(tmp, t, sizeof(tmp));
        if (!check()) vis = 0;
      }
      if (!vis) {
        swap(t1, t2);
        if (t1 == t[2]) {
          vis = 1;
          t[3] = t2;
          memcpy(tmp, t, sizeof(tmp));
          if (!check()) vis = 0;
        }
        if (!vis) {
          swap(t[2], t[1]);
          if (t1 == t[2]) {
            vis = 1;
            t[3] = t2;
            memcpy(tmp, t, sizeof(tmp));
            if (!check()) vis = 0;
          }
        }
      }
    }
    if (vis) {
      puts("YES");
      for (int i = 1; i <= n; i++) printf("%d ", tmp[i]);
      return 0;
    } else {
      t1 = 2, t2 = 1;
      if (t1 == t[2]) {
        vis = 1;
        t[3] = t2;
        memcpy(tmp, t, sizeof(tmp));
        if (!check()) vis = 0;
      }
      if (!vis) {
        swap(t[2], t[1]);
        if (t1 == t[2]) {
          vis = 1;
          t[3] = t2;
          memcpy(tmp, t, sizeof(tmp));
          if (!check()) vis = 0;
        }
        if (!vis) {
          swap(t1, t2);
          if (t1 == t[2]) {
            vis = 1;
            t[3] = t2;
            memcpy(tmp, t, sizeof(tmp));
            if (!check()) vis = 0;
          }
          if (!vis) {
            swap(t[2], t[1]);
            if (t1 == t[2]) {
              vis = 1;
              t[3] = t2;
              memcpy(tmp, t, sizeof(tmp));
              if (!check()) vis = 0;
            } else {
              puts("NO");
              return 0;
            }
          }
        }
      }
    }
  } else {
    if (t1 == t[2]) {
      vis = 1;
      t[3] = t2;
      memcpy(tmp, t, sizeof(tmp));
      if (!check()) vis = 0;
    }
    if (!vis) {
      swap(t[2], t[1]);
      if (t1 == t[2]) {
        vis = 1;
        t[3] = t2;
        memcpy(tmp, t, sizeof(tmp));
        if (!check()) vis = 0;
      }
      if (!vis) {
        swap(t1, t2);
        if (t1 == t[2]) {
          vis = 1;
          t[3] = t2;
          memcpy(tmp, t, sizeof(tmp));
          if (!check()) vis = 0;
        }
        if (!vis) {
          swap(t[2], t[1]);
          if (t1 == t[2]) {
            vis = 1;
            t[3] = t2;
            memcpy(tmp, t, sizeof(tmp));
            if (!check()) vis = 0;
          } else {
            puts("NO");
            return 0;
          }
        }
      }
    }
  }
  if (vis) puts("YES");
  for (int i = 1; i <= n; i++) printf("%d ", tmp[i]);
  return 0;
}
