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>
#include <string.h>
using namespace std;
#include <ext/pb_ds/assoc_container.hpp> //required
#include <ext/pb_ds/tree_policy.hpp> //required
using namespace __gnu_pbds; //required 
template <typename T> using ordered_set =  tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>; 

#if DEBUG && !ONLINE_JUDGE
    #include "header.h"
#else
    #define debug(args...)
#endif

typedef vector<int> vi;
typedef pair<int,int> ii;
typedef vector<ii> vii;
typedef pair<int,ii> iii;
typedef vector<pair<ii,int> > viii;
typedef vector<vi> vvi;
typedef vector<vii> vvii;
typedef map<int,int> mii;
typedef map<char,int> mci;
typedef long long lli;
typedef long double ld;
typedef vector<lli> vlli;


#define pb push_back
#define trav(a,x) for(auto& a : x)
#define all(x) x.begin(),x.end()
#define SZ(x) (int)(x).size()
#define IT iterator
#define fi first
#define se second
#define lb lower_bound
#define ub upper_bound
#define mp(a,b) make_pair(a,b)
#define init(v,n) (v).clear();(v).resize(n)
#define initi(v,n,i) (v).clear();(v).resize(n,i)
#define rep(i,a,b) for(auto i=(a);i<b;i++)

#define INF (int)1e9
#define EPS 1e-9;
#define PI 3.14159265359
#define swap(a,b) { auto c = a; a=b; b=c; }
#define EULER 2.7182818284
#define MOD 1000000007
void preprocess(void) {
    return;
}
int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.precision(20);
    //freopen("input.txt","r",stdin);
    //freopen("output.txt","w",stdout);
    preprocess();
    int n; cin>>n;
    vi a(n), b(n);
    rep(i,0,n-1) cin>>a[i];
    rep(i,0,n-1) cin>>b[i];
    int t[100001];
    rep(t0, 0, 4) {
        int flag = 0;
        t[0] = t0;
        debug(t0);
        rep(i,0,n-1) {
            if(a[i] == 0) {
                if(t[i] == 0)
                    t[i+1] = 0;
                else {
                    flag = 1;
                    break;
                }
            }
            if(a[i] == 1) {
                if(b[i] == 0 && t[i] < 2) {
                    if(t[i] == 0) t[i+1] = 1;
                    if(t[i] == 1) t[i+1] = 0;
                }
                else if(b[i] == 1 && t[i] == 1) {
                    if(t[i] == 1) t[i+1] = 1;
                }
                else {
                    flag = 1;
                    break;
                }
            }
            if(a[i] == 2) {
                if(b[i] == 0 && (t[i] == 0 || t[i] == 2) ) {
                    if(t[i] == 2) t[i+1] = 0;
                    if(t[i] == 0) t[i+1] = 2;
                }
                else if(b[i] == 2 && (t[i] == 2)) {
                    t[i+1] = 2;
                }
                else {
                    flag = 1;
                    break;
                }
            }
            if(a[i] == 3) {
                if(b[i] == 0 &&(t[i] == 1 || t[i] == 2)) {
                    t[i+1] = (t[i] == 1)?2:1;
                    debug(i,t[i], t[i+1]);
                }
                else if(b[i] == 1 &&(t[i] == 1 || t[i] == 3)) {
                    t[i+1] = (t[i] == 1)?3:1;
                }
                else if(b[i] == 2 && (t[i]>1)) {
                    t[i+1] = (t[i] == 2)?3:2;
                }
                else if(b[i] == 3 && t[i] == 3) {
                    t[i+1] = 3;
                }
                else {
                    flag = 1;
                    break;
                }
            }
        }
        if(!flag) {
            cout<<"YES"<<endl;
            rep(i,0,n) cout<<t[i]<<" ";
            return 0;
        }
    }
    cout<<"NO"<<endl;

}

