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<bool> sieve(long long n) {
  vector<bool> prime(n + 1, true);
  prime[0] = prime[1] = false;
  for (long long i = 2; i <= n; ++i) {
    if (prime[i] && (i * i) <= n)
      for (long long j = i * i; j <= n; j += i) prime[j] = false;
  }
  return prime;
}
long long power(long long a, long long b, long long m = 1000000007) {
  a %= m;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n;
  cin >> n;
  vector<pair<long long, long long> > cases{
      {0, 0}, {1, 0}, {1, 1}, {2, 0}, {2, 2}, {3, 0}, {3, 1}, {3, 2}, {3, 3}};
  vector<long long> a(n - 1), b(n - 1);
  for (long long i = 0; i < (long long)n - 1; ++i) cin >> a[i];
  for (long long i = 0; i < (long long)n - 1; ++i) cin >> b[i];
  long long dp[4][n];
  memset(dp, 0, sizeof(dp));
  for (long long i = 0; i < (long long)n - 1; ++i) {
    long long val1 = a[i];
    long long val2 = b[i];
    pair<long long, long long> temp = {val1, val2};
    bool ok = 0;
    for (long long j = 0; j < (long long)cases.size(); ++j) {
      if (cases[j] == temp) {
        ok = 1;
        break;
      }
    }
    if (!ok) {
      cout << "NO";
      return 0;
    }
    dp[val1][i] += 1;
    dp[val2][i] += 1;
    dp[val1][i + 1] += 1;
    dp[val2][i + 1] += 1;
  }
  vector<long long> res(n, 0);
  for (long long i = 1; i < n - 1; ++i) {
    long long mx = 0;
    long long val;
    for (long long j = 0; j < 4; ++j) {
      if (dp[j][i] > mx) {
        mx = dp[j][i];
        val = j;
      }
    }
    res[i] = val;
  }
  pair<long long, long long> temp = {a[0], b[0]};
  if (res[1] == temp.first)
    res[0] = temp.second;
  else
    res[0] = temp.first;
  temp = {a[n - 2], b[n - 2]};
  if (res[n - 2] == temp.first)
    res[n - 1] = temp.second;
  else
    res[n - 1] = temp.first;
  cout << "YES\n";
  for (long long i = 0; i < (long long)n; ++i) cout << res[i] << " ";
  return 0;
}
