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>
#pragma GCC optimize("O3")
using namespace std;
mt19937 rnd((uint64_t) new char);
int64_t rand(int64_t x, int64_t y) { return rnd() % (y - x + 1) + x; }
template <typename t>
t min(t a) {
  return a;
}
template <typename t, typename... ts>
t min(const t& a, const ts&... as) {
  t tmp = min(as...);
  return tmp < a ? tmp : a;
}
template <typename t>
t max(t a) {
  return a;
}
template <typename t, typename... ts>
t max(const t& a, const ts&... as) {
  t tmp = max(as...);
  return tmp > a ? tmp : a;
}
void prn(ostream& os) { os << '\n'; }
void prn() { cout << '\n'; }
template <typename t, typename... ts>
void prn(ostream& os, const t& a, const ts&... b) {
  os << a << ' ';
  prn(os, b...);
}
template <typename t, typename... ts>
void prn(const t& a, const ts&... b) {
  cout << a << ' ';
  prn(cout, b...);
}
using vi = vector<int>;
using vll = vector<int64_t>;
using pii = pair<int, int>;
using pll = pair<int64_t, int64_t>;
using ll = int64_t;
set<pair<int, int>> gg = {{1, 0}, {0, 1}, {2, 0}, {0, 2},
                          {1, 3}, {2, 3}, {0, 3}};
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  int n;
  cin >> n;
  vector<int> a(n - 1);
  for (auto& i : a) cin >> i;
  ;
  vector<int> b(n - 1);
  for (auto& i : b) cin >> i;
  ;
  vector<int> t(n);
  for (int i = 0; i < n - 1; i++) {
    t[i] |= b[i];
    t[i + 1] |= b[i];
    if ((t[i] & t[i + 1]) != b[i]) {
      cout << "NO\n";
      return 0;
    }
    if (b[i] == 0) {
      if (a[i] == 0) continue;
      if (a[i] == 1) {
        if (t[i] == 1) continue;
        if (i == n - 2) {
          t[i + 1] = a[i];
          continue;
        }
        if (a[i + 1] == 1) {
          t[i + 1] = 1;
        }
        if (a[i + 1] == 2) {
          t[i] = 1;
          t[i + 1] = 0;
        }
        if (a[i + 1] == 0) {
          if (i == 0) {
            t[i] = 1;
          } else if (a[i - 1] == 2 || a[i - 1] == 0 || a[i - 1] == 3) {
            cout << "NO\n";
            return 0;
          }
        }
        if (i != 0 && a[i - 2] == 2 && t[i] == 1) {
          cout << "NO\n";
          return 0;
        }
        if (a[i + 1] == 3) {
          cout << "NO\n";
          return 0;
        }
        if (i != 0 && a[i - 1] == 2)
          t[i + 1] = 1;
        else
          t[i] = 1;
      }
      if (a[i] == 2) {
        if (t[i] == 2) continue;
        if (i == n - 2) {
          t[i + 1] = a[i];
          continue;
        }
        if (a[i + 1] == 2) {
          t[i + 1] = 2;
        }
        if (a[i + 1] == 1) {
          t[i] = 1;
          t[i + 1] = 0;
        }
        if (a[i + 1] == 0) {
          if (i == 0) {
            t[i] = 1;
          } else if (a[i - 1] == 1 || a[i - 1] == 0 || a[i - 1] == 3) {
            cout << "NO\n";
            return 0;
          }
        }
        if (a[i + 1] == 3) {
          cout << "NO\n";
          return 0;
        }
        if (i != 0 && a[i - 2] == 1 && t[i] == 2) {
          cout << "NO\n";
          return 0;
        }
        if (i != 0 && a[i - 1] == 1)
          t[i + 1] = 2;
        else
          t[i] = 2;
      }
      if (a[i] == 3) {
        if (t[i] != 0) {
          cout << "NO\n";
          return 0;
        }
        t[i] = 0;
        t[i + 1] = 3;
      }
    } else {
      if (b[i] == 3) {
        if (a[i] != 3) {
          cout << "NO\n";
          return 0;
        }
      }
      if (b[i] == 2) {
        if (a[i] != 2 && a[i] != 3) {
          prn(a[i]);
          cout << "NO\n";
          return 0;
        }
        if (a[i] == 3 && t[i] != 3) {
          t[i + 1] = 3;
        }
        if (a[i] == 2 && t[i] == 3) {
          cout << "NO\n";
          return 0;
        }
      }
      if (b[i] == 1) {
        if (a[i] != 1 && a[i] != 3) {
          cout << "NO\n";
          return 0;
        }
        if (a[i] == 3 && t[i] != 3) {
          t[i + 1] = 3;
        }
        if (a[i] == 1 && t[i] == 3) {
          cout << "NO\n";
          return 0;
        }
      }
    }
    if (i != 0) {
      if ((t[i - 1] | t[i]) != a[i - 1]) {
        cout << "NO\n";
        return 0;
      }
    }
  }
  cout << "YES\n";
  for (const auto& i : t) cout << i << ' ';
  cout << '\n';
  ;
}
