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 int MOD = (int)1e9 + 7;
const double EPS = 1e-9;
const int INF = INT_MAX;
const long long INFLL = (long long)1e18;
const double PI = acos(-1.0);
template <class T>
T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
template <class T>
T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
const int N = (int)1e5 + 5;
const int P = 1 << 18;
int B[4][4], A[4][4];
int n;
int a[N], b[N];
int t[N];
vector<int> v[N];
int main() {
  ios::sync_with_stdio(false);
  cout.setf(ios::fixed | ios::showpoint);
  cout.precision(8);
  for (int i = 0; i < 4; i++) {
    for (int j = 0; j < 4; j++) {
      A[i][j] = i | j;
    }
  }
  for (int i = 0; i < 4; i++) {
    for (int j = 0; j < 4; j++) {
      B[i][j] = i & j;
    }
  }
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    cin >> a[i];
  }
  for (int i = 0; i < n - 1; i++) {
    cin >> b[i];
  }
  for (int i = 0; i < n - 1; i++) {
    bool flag = false;
    for (int k = 0; k < 4; k++) {
      for (int l = k; l < 4; l++) {
        if (a[i] == A[k][l] && b[i] == B[k][l]) {
          flag = true;
          v[i].push_back(k);
          v[i].push_back(l);
        }
      }
    }
    if (!flag) {
      return !printf("NO\n");
    }
  }
  bool ok = false;
  for (int x = 0; x < (int)v[0].size(); x++) {
    int cur;
    if (x % 2 == 0)
      cur = v[0][x + 1];
    else
      cur = v[0][x - 1];
    t[0] = v[0][x];
    int cnt = 1;
    for (int i = 1; i < n - 1; i++) {
      bool flag = false;
      int pos = -1;
      for (int j = 0; j < (int)v[i].size(); j++) {
        int can = v[i][j];
        if (can == cur) {
          flag = true;
          pos = j;
          break;
        }
      }
      if (flag) {
        cnt++;
        t[i] = cur;
        if (pos % 2 == 0)
          cur = v[i][pos + 1];
        else
          cur = v[i][pos - 1];
      }
    }
    if (cnt == n - 1) {
      ok = true;
      t[n - 1] = cur;
      break;
    }
  }
  if (!ok) return !printf("NO\n");
  cout << "YES\n";
  for (int i = 0; i < n; i++) {
    cout << t[i] << ' ';
  }
  cout << '\n';
  return 0;
}
