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.
#pragma GCC optimize("Ofast")
#pragma GCC optimize ("unroll-loops")
#pragma GCC optimize ("fast-math")
// #pragma GCC target("avx2")
#pragma GCC optimize("O3")
#include<bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
#define ll long long
#define int long long
#define pb push_back
#define F first
#define S second
#define ld long double
#define pii pair < int,  int >
#define tip  int
using namespace std;
const ld eps = 1e-13;
const ll inf = 1e9;
//const size_t N = 1e5+ 100;
const ll MOD = 1e9+7;


typedef tree<
tip,
null_type,
less < tip >,
rb_tree_tag,
tree_order_statistics_node_update> ordered_set;
mt19937_64 gen(time(0));
int rx[7]{0, -1, 1, 0, -1, 0, 1};
int ry[7]{-1, -1, 0, 0, 0, 1, 1};
const int N = 2e6 + 10;
int dp[N][4];
int32_t main()
{
cin.tie(0);
cout.tie(0);
    ios_base::sync_with_stdio(0);

#ifdef LOCAL
freopen("input.txt","r", stdin);
freopen("output.txt","w", stdout);
//freopen("b2.txt","r", stdin);
//freopen("output1.txt","w", stdout);
#else
#define cerr if(0)cerr

#endif // LOCAL

    ll n;
    cin >> n;
    n--;
    vector < int > a(n);
    vector < int > b(n);

    for(int i = 0; i < n; i++)
        cin >> a[i];
    for(int i = 0; i < n; i++)
        cin >> b[i];
    n++;
    vector < int > t(n);
{
    t[0] = 1;
    bool can = 1;
    for(int i = 1; i < n && can; i++){
        bool f = 0;
        for(int j = 1; j<= 3; j++){
            t[i]= j;
            if(t[i-1] | t[i] == a[i - 1])
                if(t[i-1] & t[i] == b[i - 1])
                    {

                        f= 1;
                        break;
                    }
        }
    if(!f)
        can = 0;
    }
if(can){
                cout <<"YES\n";

    for(auto to : t)
        cout << to << " ";
    return 0;
}
}

{
    t[0] = 2;
    bool can = 1;
    for(int i = 1; i < n && can; i++){
        bool f = 0;
        for(int j = 1; j<= 3; j++){
            t[i]= j;
            if(t[i-1] | t[i] == a[i - 1])
                if(t[i-1] & t[i] == b[i - 1])
                    {

                        f= 1;
                        break;
                    }
        }
    if(!f)
        can = 0;
    }
if(can){
                cout <<"YES\n";

    for(auto to : t)
        cout << to << " ";
    return 0;
}
}
{
    t[0] = 3;
    bool can = 1;
    for(int i = 1; i < n && can; i++){
        bool f = 0;
        for(int j = 1; j<= 3; j++){
            t[i]= j;
            if(t[i-1] | t[i] == a[i - 1])
                if(t[i-1] & t[i] == b[i - 1])
                    {

                        f= 1;
                        break;
                    }
        }
    if(!f)
        can = 0;
    }
if(can){
        cout <<"YES\n";
    for(auto to : t)
        cout << to << " ";
    return 0;
}
}
cout <<"NO\n";
}
