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;
long long n, m, k;
vector<long long> A, C;
vector<tuple<long long, long long, long long, long long>> Used;
void run(long long num) {
  vector<long long> Ans;
  Ans.push_back(num);
  for (long long x = n - 2; x >= 0; x--) {
    if (num == 0) {
      num = A[x];
    } else if (num == 1) {
      if (C[x] == 0 && A[x] == 1) {
        num = 0;
      } else if (C[x] == 0 && A[x] == 3) {
        num = 2;
      } else if (C[x] == 1 && A[x] == 1) {
        num = 1;
      } else if (C[x] == 1 && A[x] == 3) {
        num = 3;
      } else {
        return;
      }
    } else if (num == 2) {
      if (C[x] == 0 && A[x] == 2) {
        num = 0;
      } else if (C[x] == 0 && A[x] == 3) {
        num = 1;
      } else if (C[x] == 2 && A[x] == 2) {
        num = 2;
      } else if (C[x] == 2 && A[x] == 3) {
        num = 3;
      } else {
        return;
      }
    } else if (num == 3) {
      if (C[x] == 0 && A[x] == 3) {
        num = 0;
      } else if (C[x] == 1 && A[x] == 3) {
        num = 1;
      } else if (C[x] == 2 && A[x] == 3) {
        num = 2;
      } else if (C[x] == 3 && A[x] == 3) {
        num = 3;
      } else {
        return;
      }
    }
    Ans.push_back(num);
  }
  reverse(Ans.begin(), Ans.end());
  cout << "YES\n";
  for (long long x : Ans) cout << x << " ";
  exit(0);
}
void dfs(long long x, long long num) {
  if (x == n - 1) run(num);
  if (num == 0) {
    if (get<0>(Used[x])) return;
    get<0>(Used[x]) = 1;
  } else if (num == 1) {
    if (get<1>(Used[x])) return;
    get<1>(Used[x]) = 1;
  } else if (num == 2) {
    if (get<2>(Used[x])) return;
    get<2>(Used[x]) = 1;
  } else if (num == 3) {
    if (get<3>(Used[x])) return;
    get<3>(Used[x]) = 1;
  }
  if (num == 0) {
    if (C[x] != 0) return;
    dfs(x + 1, A[x]);
  } else if (num == 1) {
    if (C[x] == 0 && A[x] == 1) {
      dfs(x + 1, 0);
    } else if (C[x] == 0 && A[x] == 3) {
      dfs(x + 1, 2);
    } else if (C[x] == 1 && A[x] == 1) {
      dfs(x + 1, 1);
    } else if (C[x] == 1 && A[x] == 3) {
      dfs(x + 1, 3);
    } else {
      return;
    }
  } else if (num == 2) {
    if (C[x] == 0 && A[x] == 2) {
      dfs(x + 1, 0);
    } else if (C[x] == 0 && A[x] == 3) {
      dfs(x + 1, 1);
    } else if (C[x] == 2 && A[x] == 2) {
      dfs(x + 1, 2);
    } else if (C[x] == 2 && A[x] == 3) {
      dfs(x + 1, 3);
    } else {
      return;
    }
  } else if (num == 3) {
    if (C[x] == 0 && A[x] == 3) {
      dfs(x + 1, 0);
    } else if (C[x] == 1 && A[x] == 3) {
      dfs(x + 1, 1);
    } else if (C[x] == 2 && A[x] == 3) {
      dfs(x + 1, 2);
    } else if (C[x] == 3 && A[x] == 3) {
      dfs(x + 1, 3);
    } else {
      return;
    }
  }
}
signed main() {
  cin >> n;
  A.resize(n - 1);
  C.resize(n - 1);
  for (long long &x : A) cin >> x;
  for (long long &x : C) cin >> x;
  Used.assign(n, make_tuple(0, 0, 0, 0));
  dfs(0, 0);
  dfs(0, 1);
  dfs(0, 2);
  dfs(0, 3);
  cout << "NO";
}
