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.
///Foxi_Doree
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

#define pb push_back
#define gcd __gcd
#define time { ios::sync_with_stdio(0); ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0); srand(time(0)); }

using namespace std;
using namespace __gnu_pbds;

ifstream in("input.txt");
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
typedef pair < ll, ll > pll;
typedef pair < double, double > pdd;
typedef vector < ll > vll;
typedef set < ll > sll;
/*
    find_by_order()
    order_of_key()
*/
typedef tree <
ll, null_type,
less < ll >, rb_tree_tag,
tree_order_statistics_node_update > set_t;
const ll inf = 1000000007, oo = 1e6 + 2, o = 4e5 + 2, f = 998244353;

int main()
{
    time

    ll n, i;

    vll a(o), b(o), ans(o);

    cin >> n;

    for(i = 0; i < n - 1; ++i) {
        cin >> a[i];
    }
    for(i = 0; i < n - 1; ++i) {
        cin >> b[i];
    }

    for(i = 0; i < n - 1; ++i) {
        if(a[i] < b[i] || (a[i] == 2 && b[i] == 1)) return cout << "NO", 0;
    }

    for(i = 0; i < n - 1; ++i) {
        if(a[i] == 3 && b[i] == 0) {
            a[i] = 2;
            b[i] = 1;
        }
    }

    if(n == 2) {
        cout << "YES" << endl;
        cout << a[0] << ' ' << b[0];
        return 0;
    }

    for(i = 0; i < n - 2; ++i) {
        if(a[i] == a[i + 1] || a[i] == b[i + 1]) {
            ans[i] = b[i];
            ans[i + 1] = a[i];
        }
        else {
            if(b[i] == b[i + 1] || b[i] == a[i + 1]) {
                ans[i] = a[i];
                ans[i + 1] = b[i];
            }
            else {
                return cout << "NO", 0;
            }
        }
    }
    cout << "YES" << endl;
    for(i = 0; i < n - 1; ++i) {
        cout << ans[i] << ' ';
    }
    if(ans[n - 2] == a[n - 2]) cout << b[n - 2];
    else cout << a[n - 2];
}
