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;
const int INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3f;
const int MOD = 1e9 + 7;
const int N = 1e6 + 7;
int n;
vector<int> a, b;
int aa[N], bb[N], t[N << 1];
void Init() {
  memset(t, -1, sizeof(t));
  a.clear();
  b.clear();
  for (int i = (1); i < (n); ++i) {
    scanf("%d", &aa[i - 1]);
    for (int j = (2) - 1; j >= (0); --j) a.push_back((aa[i - 1] >> j) & 1);
  }
  for (int i = (1); i < (n); ++i) {
    scanf("%d", &bb[i - 1]);
    for (int j = (2) - 1; j >= (0); --j) b.push_back((bb[i - 1] >> j) & 1);
  }
}
int Solve() {
  for (int i = (0); i < ((int)(a).size()); ++i) {
    if (a[i] == 0 && b[i] == 1) return puts("NO");
    if (!a[i]) {
      if (t[i] != 2 && t[i] != -1 && t[i] != 0) return puts("NO");
      t[i] = t[i + 2] = 0;
    } else if (b[i]) {
      if (t[i] != 2 && t[i] != -1 && t[i] != 1) return puts("NO");
      t[i] = t[i + 2] = 1;
    } else {
      if (t[i] == -1)
        t[i] = t[i + 2] = 2;
      else if (t[i] == 0)
        t[i + 2] = 1;
      else if (t[i] == 1)
        t[i + 2] = 0;
      else
        t[i + 2] = 2;
    }
  }
  for (int i = (0); i < ((int)(a).size() + 2); ++i)
    if (t[i] != 2) {
      int pos = i - 2;
      while (pos >= 0 && t[pos] == 2) {
        t[pos] = 1 - t[pos + 2];
        pos -= 2;
      }
      pos = i + 2;
      while (pos < (int)(a).size() + 2 && t[pos] == 2) {
        t[pos] = 1 - t[pos - 2];
        pos += 2;
      }
    }
  for (int i = (0); i < (2); ++i) {
    int pos;
    if (t[i] == 2) {
      t[pos = i] = 0;
      while (pos + 2 < (int)(a).size() + 2) {
        t[pos + 2] = 1 - t[pos];
        pos += 2;
      }
    }
  }
  vector<int> ans;
  for (int i = 0; i < (int)(a).size() + 2; i += 2)
    ans.push_back(t[i] << 1 | t[i + 1]);
  for (int i = (0); i < ((int)(ans).size() - 1); ++i) {
    int tmp = ans[i] | ans[i + 1];
    if (tmp != aa[i]) return puts("NO");
    tmp = ans[i] & ans[i + 1];
    if (tmp != bb[i]) return puts("NO");
  }
  puts("YES");
  for (int i = (0); i < ((int)(ans).size()); ++i)
    printf("%d%c", ans[i], " \n"[i == (int)(ans).size() - 1]);
  return 0;
}
int main() {
  while (~scanf("%d", &n)) {
    Init();
    Solve();
  }
  return 0;
}
