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;
double PI = acos(-1.0);
long long mod = 1e9 + 7;
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
long long GCD(long long a, long long b) {
  if (b == 0) return a;
  return GCD(b, a % b);
}
long long prod(long long a, long long b) {
  long long res = 0;
  a %= mod;
  while (b) {
    if (b & 1ll) res = (res + a) % mod;
    a = (2ll * a) % mod;
    b >>= 1ll;
  }
  return res;
}
long long power(long long a, long long b) {
  long long res = 1ll;
  while (b > 0) {
    if (b % 2ll) res = (res * a) % mod;
    a = (a * a) % mod;
    b /= 2ll;
  }
  return res;
}
long long n;
vector<long long> getArray(vector<long long> &a, vector<long long> &b) {
  bool f = true;
  vector<long long> t(n);
  t[0] = 1;
  for (long long i = 0; i < n - 1; i++) {
    if (t[i] == 1) {
      if (a[i] == 1 and b[i] == 1) {
        t[i + 1] = 1;
      } else if (a[i] == 0 and b[i] == 1) {
        f = false;
        break;
      } else if (a[i] == 1 and b[i] == 0) {
        t[i + 1] = 0;
      } else {
        f = false;
        break;
      }
    } else {
      if (a[i] == 1 and b[i] == 1) {
        f = false;
        break;
      } else if (a[i] == 0 and b[i] == 1) {
        f = false;
        break;
      } else if (a[i] == 1 and b[i] == 0) {
        t[i + 1] = 1;
      } else {
        t[i + 1] = 0;
      }
    }
  }
  if (f) {
    return t;
  }
  f = true;
  t[0] = 0;
  for (long long i = 0; i < n - 1; i++) {
    if (t[i] == 1) {
      if (a[i] == 1 and b[i] == 1) {
        t[i + 1] = 1;
      } else if (a[i] == 0 and b[i] == 1) {
        f = false;
        break;
      } else if (a[i] == 1 and b[i] == 0) {
        t[i + 1] = 0;
      } else {
        f = false;
        break;
      }
    } else {
      if (a[i] == 1 and b[i] == 1) {
        f = false;
        break;
      } else if (a[i] == 0 and b[i] == 1) {
        f = false;
        break;
      } else if (a[i] == 1 and b[i] == 0) {
        t[i + 1] = 1;
      } else {
        t[i + 1] = 0;
      }
    }
  }
  if (f) return t;
  return {};
}
void solve() {
  cin >> n;
  vector<long long> a1(n - 1), a2(n - 1), b1(n - 1), b2(n - 1);
  for (long long i = 0; i < n - 1; i++) {
    long long x;
    cin >> x;
    a1[i] = 1 & x;
    a2[i] = 2 & x;
    a2[i] >>= 1;
  }
  for (long long i = 0; i < n - 1; i++) {
    long long x;
    cin >> x;
    b1[i] = 1 & x;
    b2[i] = 2 & x;
    b2[i] >>= 1;
  }
  vector<long long> t(n, 3);
  vector<long long> f = getArray(a1, b1);
  vector<long long> g = getArray(a2, b2);
  if (f.size() == 0 || g.size() == 0) {
    cout << "NO"
         << "\n";
    return;
  }
  cout << "YES"
       << "\n";
  for (long long i = 0; i < n; i++) {
    g[i] <<= 1;
    f[i] |= g[i];
    cout << f[i] << " ";
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t = 1;
  while (t--) {
    solve();
  }
}
