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;
const long long mod = 1e9 + 7;
const long long INF = 9e18;
const int inf = 2147483647;
long long phi(long long n) {
  long long ans = n;
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      ans = ans / i * (i - 1);
      while (n % i == 0) n /= i;
    }
  }
  if (n > 1) ans = ans / n * (n - 1);
  return ans;
}
long long qpow(long long a, long long b, long long mod) {
  long long r = 1;
  for (; b; b >>= 1) {
    if (b & 1) r = (r * a) % mod;
    a = (a * a) % mod;
  }
  return r;
}
long long C(int n, int m) {
  if (m < n - m) m = n - m;
  long long ans = 1;
  for (int i = m + 1; i <= n; i++) ans *= i;
  for (int i = 1; i <= n - m; i++) ans /= i;
  return ans;
}
const int N = 1e5 + 5;
int a[N], b[N], t[N];
void work() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i < n; i++) scanf("%d", &a[i]);
  for (int i = 1; i < n; i++) scanf("%d", &b[i]);
  int T = a[1] + b[1], ko = 0;
  for (int t1 = 0; t1 <= 3; t1++) {
    int t2 = T - t1;
    if (t2 < 0) break;
    int ok = 1;
    t[1] = t1, t[2] = t2;
    if ((t[1] | t[2]) != a[1]) continue;
    if ((t[1] & t[2]) != b[1]) continue;
    for (int i = 3; i <= n; i++) {
      t[i] = a[i - 1] + b[i - 1] - t[i - 1];
      if (t[i] < 0 || t[i] > 3) {
        ok = 0;
        break;
      }
      if (a[i - 1] != (t[i] | t[i - 1]) || b[i - 1] != (t[i] & t[i - 1])) {
        ok = 0;
        break;
      }
    }
    if (ok) {
      ko = 1;
      break;
    }
  }
  if (ko) {
    puts("YES");
    for (int i = 1; i <= n; i++) printf("%d ", t[i]);
  } else
    puts("NO");
}
int main() {
  work();
  return 0;
}
