#include <bits/stdc++.h>
const long long MXN = 1e5 + 1;
const long long MNN = 2e3 + 1;
const long long MOD = 1e9 + 7;
const long long INF = 1e18;
const long long OO = 1e9 + 500;
using namespace std;
int n, a[MXN], b[MXN], ans[MXN];
vector<pair<int, int> > q[4][4];
bool dfs(int x, int prb) {
  if (x == n) {
    ans[x] = prb;
    return true;
  }
  for (int j = 0; j < q[a[x]][b[x]].size(); j++) {
    int tol = q[a[x]][b[x]][j].first, tor = q[a[x]][b[x]][j].second;
    if (prb == tol) {
      if (dfs(x + 1, tor)) {
        ans[x] = tol;
        return true;
      }
    }
  }
  return false;
}
int main() {
  ios_base::sync_with_stdio(0), cout.tie(0), cin.tie(0);
  ;
  cin >> n;
  for (long long i = 1; i <= n - 1; i++) {
    cin >> a[i];
  }
  for (long long i = 1; i <= n - 1; i++) {
    cin >> b[i];
  }
  for (long long i = 1; i <= n - 1; i++) {
    if (a[i] < b[i]) {
      cout << "NO";
      exit(0);
    }
    if (a[i] == 2 && b[i] == 1) {
      continue;
    }
    q[a[i]][b[i]].push_back(make_pair(a[i], b[i]));
    q[a[i]][b[i]].push_back(make_pair(b[i], a[i]));
    if (a[i] == 3 && b[i] == 0) {
      q[a[i]][b[i]].push_back(make_pair(2, 1));
      q[a[i]][b[i]].push_back(make_pair(1, 2));
    }
  }
  for (int i = 0; i < q[a[1]][b[1]].size(); i++) {
    int tr = q[a[1]][b[1]][i].second;
    if (dfs(2, tr)) {
      cout << "YES" << '\n';
      cout << q[a[1]][b[1]][i].first << ' ';
      for (int j = 2; j <= n; j++) {
        cout << ans[j] << ' ';
      }
      exit(0);
    }
  }
  cout << "NO";
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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.

### Response

```cpp
#include <bits/stdc++.h>
const long long MXN = 1e5 + 1;
const long long MNN = 2e3 + 1;
const long long MOD = 1e9 + 7;
const long long INF = 1e18;
const long long OO = 1e9 + 500;
using namespace std;
int n, a[MXN], b[MXN], ans[MXN];
vector<pair<int, int> > q[4][4];
bool dfs(int x, int prb) {
  if (x == n) {
    ans[x] = prb;
    return true;
  }
  for (int j = 0; j < q[a[x]][b[x]].size(); j++) {
    int tol = q[a[x]][b[x]][j].first, tor = q[a[x]][b[x]][j].second;
    if (prb == tol) {
      if (dfs(x + 1, tor)) {
        ans[x] = tol;
        return true;
      }
    }
  }
  return false;
}
int main() {
  ios_base::sync_with_stdio(0), cout.tie(0), cin.tie(0);
  ;
  cin >> n;
  for (long long i = 1; i <= n - 1; i++) {
    cin >> a[i];
  }
  for (long long i = 1; i <= n - 1; i++) {
    cin >> b[i];
  }
  for (long long i = 1; i <= n - 1; i++) {
    if (a[i] < b[i]) {
      cout << "NO";
      exit(0);
    }
    if (a[i] == 2 && b[i] == 1) {
      continue;
    }
    q[a[i]][b[i]].push_back(make_pair(a[i], b[i]));
    q[a[i]][b[i]].push_back(make_pair(b[i], a[i]));
    if (a[i] == 3 && b[i] == 0) {
      q[a[i]][b[i]].push_back(make_pair(2, 1));
      q[a[i]][b[i]].push_back(make_pair(1, 2));
    }
  }
  for (int i = 0; i < q[a[1]][b[1]].size(); i++) {
    int tr = q[a[1]][b[1]][i].second;
    if (dfs(2, tr)) {
      cout << "YES" << '\n';
      cout << q[a[1]][b[1]][i].first << ' ';
      for (int j = 2; j <= n; j++) {
        cout << ans[j] << ' ';
      }
      exit(0);
    }
  }
  cout << "NO";
  return 0;
}

```