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>
const long long MOD = 1e9 + 7;
const long long MAXN = 1e6 + 1;
using namespace std;
long long readInt() {
  bool minus1 = false;
  long long result = 0;
  char ch;
  ch = getchar();
  while (true) {
    if (ch == '-') break;
    if (ch >= '0' && ch <= '9') break;
    ch = getchar();
  }
  if (ch == '-')
    minus1 = true;
  else
    result = ch - '0';
  while (true) {
    ch = getchar();
    if (ch < '0' || ch > '9') break;
    result = result * 10 + (ch - '0');
  }
  if (minus1)
    return -result;
  else
    return result;
}
int a[MAXN];
int b[MAXN];
int n;
int M[4][4];
vector<pair<int, int> > var[5][5];
pair<int, int> first(vector<pair<int, int> >& v, int x) {
  for (pair<int, int> C : v) {
    if (C.first == x) {
      return C;
    }
  }
  return {-1, -1};
}
void check(int x, int y) {
  vector<int> ans;
  ans.push_back(x);
  for (int i = 2; i < n; i++) {
    ans.push_back(y);
    pair<int, int> kek = first(var[a[i]][b[i]], y);
    if (kek.first != -1) {
      x = kek.first;
      y = kek.second;
    } else {
      return;
    }
  }
  ans.push_back(y);
  cout << "YES\n";
  for (int i : ans) {
    cout << i << ' ';
  }
  exit(0);
}
int main() {
  for (int i = 0; i <= 3; i++) {
    for (int j = 0; j <= 3; j++) {
      M[(i | j)][(i & j)] = 1;
    }
  }
  n = readInt();
  for (int i = 1; i < n; i++) {
    a[i] = readInt();
  }
  for (int i = 1; i < n; i++) {
    b[i] = readInt();
  }
  for (int i = 1; i < n; i++) {
    if (!M[a[i]][b[i]]) {
      cout << "NO";
      return 0;
    }
  }
  for (int i = 0; i <= 3; i++) {
    for (int j = 0; j <= 3; j++) {
      var[(i | j)][(i & j)].push_back({i, j});
    }
  }
  for (pair<int, int> C : var[a[1]][b[1]]) {
    check(C.first, C.second);
    check(C.second, C.first);
  }
  return 0;
}
