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;
int MOD(int a, int b) {
  if (a > b)
    return a - b;
  else
    return b - a;
}
long long int max3(long long int a, long long int b, long long int c) {
  return max(c, max(a, b));
}
long long int power(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long int max(long long int a, long long int b) { return (a > b) ? a : b; }
long long int min(long long int a, long long int b) { return (a < b) ? a : b; }
long long int gcd(long long int a, long long int b) {
  if (b == 1)
    return 1;
  else if (a % b == 0)
    return b;
  else
    return gcd(b, a % b);
}
long long int mod(long long int x) {
  if (x >= 0)
    return x;
  else
    return -x;
}
struct objec {
  string s;
  int num;
};
int maxdist(char c) { return max(c - 'a', 'z' - c); }
char maxc(char c) { return (c > 'm') ? 'a' : 'z'; }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  int a[n - 1], b[n - 1], a1[n - 1], a2[n - 1], b1[n - 1], b2[n - 1];
  for (int i = 0; i < n - 1; i++) {
    cin >> a[i];
    if (a[i] % 2 == 0)
      a1[i] = 0;
    else
      a1[i] = 1;
    if ((a[i] / 2) % 2 == 0)
      a2[i] = 0;
    else
      a2[i] = 1;
  }
  for (int i = 0; i < n - 1; i++) {
    cin >> b[i];
    if (b[i] % 2 == 0)
      b1[i] = 0;
    else
      b1[i] = 1;
    if ((b[i] / 2) % 2 == 0)
      b2[i] = 0;
    else
      b2[i] = 1;
  }
  int c1[n], c2[n];
  int firstdef;
  for (int i = 0; i < n - 1; i++) {
    if ((a1[i] ^ b1[i]) == 0) {
      firstdef = i;
      break;
    }
  }
  if (firstdef == n) {
    c1[0] = 1;
    for (int i = 1; i < n; i++) {
      c1[i] = 1 - c1[i - 1];
    }
  } else {
    c1[firstdef] = a1[firstdef];
    for (int i = firstdef; i < n; ++i) {
      if ((a1[i] ^ b1[i]) == 0) {
        if (c1[i] != a1[i]) {
          cout << "NO";
          return 0;
        }
        c1[i + 1] = a1[i];
      } else if (a1[i] == 0 && b1[i] == 1) {
        cout << "NO";
        return 0;
      } else {
        c1[i + 1] = 1 - c1[i];
      }
    }
    for (int i = firstdef - 1; i >= 0; i--) {
      if ((a1[i] ^ b1[i]) == 0) {
        if (c1[i + 1] != a1[i]) {
          cout << "NO";
          return 0;
        }
        c1[i] = a1[i];
      } else if (a1[i] == 0 && b1[i] == 1) {
        cout << "NO";
        return 0;
      } else {
        c1[i] = 1 - c1[i + 1];
      }
    }
  }
  int firstdef2;
  for (int i = 0; i < n - 1; i++) {
    if ((a2[i] ^ b2[i]) == 0) {
      firstdef2 = i;
      break;
    }
  }
  if (firstdef2 == n) {
    c2[0] = 1;
    for (int i = 1; i < n; i++) {
      c2[i] = 1 - c2[i - 1];
    }
  } else {
    c2[firstdef2] = a2[firstdef2];
    for (int i = firstdef2; i < n; ++i) {
      if ((a2[i] ^ b2[i]) == 0) {
        if (c2[i] != a2[i]) {
          cout << "NO";
          return 0;
        }
        c2[i + 1] = a2[i];
      } else if (a2[i] == 0 && b2[i] == 1) {
        cout << "NO";
        return 0;
      } else {
        c2[i + 1] = 1 - c2[i];
      }
    }
    for (int i = firstdef2 - 1; i >= 0; i--) {
      if ((a2[i] ^ b2[i]) == 0) {
        if (c2[i + 1] != a2[i]) {
          cout << "NO";
          return 0;
        }
        c2[i] = a2[i];
      } else if (a2[i] == 0 && b2[i] == 1) {
        cout << "NO";
        return 0;
      } else {
        c2[i] = 1 - c2[i + 1];
      }
    }
  }
  cout << "YES" << endl;
  for (int i = 0; i < n; i++) {
    cout << 2 * c2[i] + c1[i] << " ";
  }
  return 0;
}
