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 int mod = (long long int)(998244353);
const long long int mod2 = (long long int)1e9 + 7;
long long int test = 1;
vector<long long int> read(int n) {
  vector<long long int> v(n);
  for (int i = 0; i < n; i++) cin >> v[i];
  return v;
}
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);
  }
};
std::vector<long long int> a;
std::vector<long long int> b;
std::vector<long long int> v;
bool go(long long int st, long long int idx, long long int n) {
  if (idx == n) {
    cout << "YES\n";
    for (auto it : v) {
      cout << it << " ";
    }
    cout << "\n";
    exit(0);
    return 1;
  }
  long long int s = a[idx - 1];
  long long int e = b[idx - 1];
  bool ok = 0;
  for (long long int i = 0; i <= 3; i++) {
    if ((i | st) == s && (i & st) == e) {
      v.push_back(i);
      ok = go(i, idx + 1, n);
      v.pop_back();
    }
  }
  return ok;
}
void solve() {
  long long int i, j, k, m, n, cnt = 0, ans = 0, sum = 0;
  cin >> n;
  a.clear();
  a.resize(n - 1);
  b.clear();
  b.resize(n - 1);
  for (auto &it : a) cin >> it;
  for (auto &it : b) cin >> it;
  bool ok = 0;
  for (long long int i = 0; i <= 3; i++) {
    v.push_back(i);
    ok = go(i, 1, n);
    v.clear();
    if (ok) {
      return;
    }
  }
  cout << "NO\n";
  test++;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int i = 1;
  solve();
  return 0;
}
