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>
const int N = (int)2e5 + 7;
const int N_ = (int)2e3 + 1;
const int MAXN = 2016 * 2016;
const int MOD = (int)1e9;
const int INF = (int)2e9;
const int DEL = (int)3e5;
const int LOG = 20;
const int SIZE = 790;
const int ALPH = 256;
const int DIGIT = 10;
const int BLOCK = 300;
const int LIGHT = 650;
const long long LINF = (int64_t)2e18;
const long double PI = 3.1415926535897932384626433832795;
const long double EPS = 1e-2;
using namespace std;
template <typename T>
T sqr(T a) {
  return a * a;
}
template <typename T>
bool uax(T &a, const T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <typename T>
bool uin(T &a, const T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
template <typename T1, typename T2>
istream &operator>>(istream &in, pair<T1, T2> &p) {
  in >> p.first >> p.second;
  return in;
}
template <typename T1, typename T2>
ostream &operator<<(ostream &out, pair<T1, T2> &p) {
  out << p.first << ' ' << p.second;
  return out;
}
int n, a[N], b[N], t[N];
map<pair<int, pair<int, int>>, int> need;
void read() {
  cin >> n;
  for (int i = 0; i < n - 1; i++) cin >> a[i];
  for (int i = 0; i < n - 1; i++) cin >> b[i];
}
void precalc() {
  for (int i = 0; i < 4; i++) {
    for (int j = 0; j < 4; j++) {
      for (int k = 0; k < 4; k++) {
        for (int g = 0; g < 4; g++) {
          if ((i | g) == j && (i & g) == k) {
            need[make_pair(i, make_pair(j, k))] = g;
          }
        }
      }
    }
  }
}
void get_try() {
  for (int i = 1; i < n; i++) {
    if (need.count(make_pair(t[i - 1], make_pair(a[i - 1], b[i - 1]))) == 0)
      return;
    t[i] = need[make_pair(t[i - 1], make_pair(a[i - 1], b[i - 1]))];
  }
  cout << "Yes\n";
  for (int i = 0; i < n; i++) cout << t[i] << ' ';
  exit(0);
}
void solve() {
  precalc();
  for (int i = 0; i < 3; i++) {
    t[0] = i;
    get_try();
  }
  cout << "No";
}
void check_time() {
  double start = clock();
  read(), solve();
  cout << "\nTime is " << (clock() - start) * 1.0 / CLOCKS_PER_SEC << "s\n";
}
int main() {
  cout << fixed << setprecision(20);
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  read();
  solve();
}
