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 long long INF = 1e18;
const long long SIZE = 1e5 + 5;
const long long mod = 1e9 + 7;
long long a[SIZE], b[SIZE];
long long dp[SIZE][5];
long long dir[SIZE][5];
long long solve(long long pos, long long prv) {
  if (pos == 0) return 1;
  if (dp[pos][prv] != -1) return dp[pos][prv];
  long long i, x, y, p(0);
  for (i = 0; i < 4; i++) {
    x = prv | i;
    y = prv & i;
    if (x == a[pos] && y == b[pos]) {
      p = solve(pos - 1, i);
      if (p) {
        dir[pos][prv] = i;
        break;
      }
    }
  }
  return dp[pos][prv] = p;
}
int main() {
  long long t = 0, n, x, y, z, i, j, k, g, p, q, ans = 0, sum = 0, c = 0;
  string s, s1, s2;
  scanf("%lld", &n);
  for (i = 1; i <= n - 1; i++) scanf("%lld", &a[i]);
  for (i = 1; i <= n - 1; i++) scanf("%lld", &b[i]);
  memset(dp, -1, sizeof dp);
  memset(dir, -1, sizeof dir);
  for (i = 0; i < 4; i++) {
    ans = solve(n - 1, i);
    if (ans) {
      p = i;
      break;
    }
  }
  if (!ans)
    puts("NO");
  else {
    vector<long long> v;
    v.push_back(p);
    x = n - 1;
    while (dir[x][p] != -1) {
      p = dir[x][p];
      x--;
      v.push_back(p);
    }
    puts("YES");
    for (i = n - 1; i >= 0; i--) cout << v[i] << " ";
    printf("\n");
  }
  return 0;
}
