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;
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
unsigned long long bpow(unsigned long long a, int n) {
  return n ? n & 1 ? bpow(a, n - 1) * a : bpow(a * a, n >> 1) : 1;
}
bool is_upper(char c) { return (c >= 65 && c <= 90); }
bool is_lower(char c) { return (c >= 97 && c <= 122); }
int get_digit(long long num, int len, int index) {
  return num / bpow(10, len - index - 1) % 10;
}
long long digit_sum(long long num) {
  long long dig_sum = 0;
  while (num > 0) {
    dig_sum += num % 10;
    num /= 10;
  }
  return dig_sum;
}
const int INF = 1e9 + 7;
const double pi = 2 * acos(0.0);
long long n, m, k, p, q, sum = 0, cnt = 0;
long long mx = INT64_MIN;
long long mn = INT64_MAX;
bool flag = 0;
vector<long long> v;
string s = "", s1, s2;
int rec(int a[], int b[], int num, int d) {
  long long temp = 4;
  long long temp1 = 4;
  long long li = 4;
  for (long long i = 0; i < 4; ++i) {
    if ((num | i) == a[d] && (num & i) == b[d]) {
      if (d == n - 2) {
        return i;
      } else {
        temp1 = rec(a, b, i, d + 1);
        if (temp1 != 4) {
          li = i;
          temp = temp1;
          break;
        }
      }
    }
  }
  if (temp != 4) v.push_back(temp);
  return li;
}
void solve() {
  cin >> n;
  int a[n - 1];
  int b[n - 1];
  for (long long i = 0; i < n - 1; ++i) {
    cin >> a[i];
  }
  for (long long i = 0; i < n - 1; ++i) {
    cin >> b[i];
  }
  long long temp = 4;
  long long temp1 = 4;
  long long li = 4;
  for (long long i = 0; i < 4; ++i) {
    temp1 = rec(a, b, i, 0);
    if (temp1 != 4) {
      li = i;
      temp = temp1;
      break;
    }
  }
  if (temp == 4) {
    cout << "NO";
  } else {
    cout << "YES" << endl;
    v.push_back(temp);
    v.push_back(li);
    reverse(v.begin(), v.end());
    {
      long long length_of_vector = v.size();
      for (long long i = 0; i < length_of_vector; ++i) {
        cout << v[i] << ' ';
      }
    };
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  solve();
  return 0;
}
