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;
;
const long long maxn = 1e6;
vector<long long> ANS1;
vector<long long> ANS2;
vector<long long> ANS3;
vector<long long> ANS4;
long long A[maxn];
long long B[maxn];
int main() {
  long long q, w, e, r, n, m, k, a, b;
  string s;
  cin >> n;
  for (q = 0; q < n - 1; q++) {
    cin >> A[q];
  }
  for (q = 0; q < n - 1; q++) {
    cin >> B[q];
  }
  bool f1 = 0, f2 = 0, f3 = 0, f4 = 0;
  ANS1.push_back(0);
  ANS2.push_back(1);
  ANS3.push_back(2);
  ANS4.push_back(3);
  for (q = 0; q < n - 1; q++) {
    if ((A[q] | B[q]) != A[q]) {
      cout << "NO";
      return 0;
    }
    if (ANS1[ANS1.size() - 1] == (A[q] & B[q])) {
      ANS1.push_back(A[q] | B[q]);
    } else {
      if (ANS1[ANS1.size() - 1] == (A[q] | B[q])) {
        ANS1.push_back(A[q] & B[q]);
      } else {
        if (A[q] == 3 && B[q] == 0) {
          ANS1.push_back(3 - ANS1[ANS1.size() - 1]);
        } else {
          f1 = 1;
        }
      }
    }
    if (ANS3[ANS3.size() - 1] == (A[q] & B[q])) {
      ANS3.push_back(A[q] | B[q]);
    } else {
      if (ANS3[ANS3.size() - 1] == (A[q] | B[q])) {
        ANS3.push_back(A[q] & B[q]);
      } else {
        if (A[q] == 3 && B[q] == 0) {
          ANS3.push_back(3 - ANS3[ANS3.size() - 1]);
        } else {
          f3 = 1;
        }
      }
    }
    if (ANS4[ANS4.size() - 1] == (A[q] & B[q])) {
      ANS4.push_back(A[q] | B[q]);
    } else {
      if (ANS4[ANS4.size() - 1] == (A[q] | B[q])) {
        ANS4.push_back(A[q] & B[q]);
      } else {
        if (A[q] == 3 && B[q] == 0) {
          ANS4.push_back(3 - ANS4[ANS4.size() - 1]);
        } else {
          f4 = 1;
        }
      }
    }
    if (ANS2[ANS2.size() - 1] == (A[q] & B[q])) {
      ANS2.push_back(A[q] | B[q]);
    } else {
      if (ANS2[ANS2.size() - 1] == (A[q] | B[q])) {
        ANS2.push_back(A[q] & B[q]);
      } else {
        if (A[q] == 3 && B[q] == 0) {
          ANS2.push_back(3 - ANS2[ANS2.size() - 1]);
        } else {
          f2 = 1;
        }
      }
    }
  }
  if (f1 && f2 && f3 && f4) {
    cout << "NO";
    return 0;
  }
  cout << "YES" << endl;
  if (!f3) {
    for (q = 0; q < ANS3.size(); q++) {
      cout << ANS3[q] << " ";
    }
    return 0;
  }
  if (!f4) {
    for (q = 0; q < ANS4.size(); q++) {
      cout << ANS4[q] << " ";
    }
    return 0;
  }
  if (!f1) {
    for (q = 0; q < ANS1.size(); q++) {
      cout << ANS1[q] << " ";
    }
  } else {
    for (q = 0; q < ANS2.size(); q++) {
      cout << ANS2[q] << " ";
    }
  }
  return 0;
}
