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 <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define ll long long
#define ld long double
#define fi first
#define se second
#define pb push_back
#define all(v) v.begin(),v.end()
#define repf(i,a,b) for(ll i=a;i<b;i++)
#define repb(i,a,b) for(ll i=a;i>=b;i--)
#define IOS ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define mod 1000000007
#define MOD 998244353 

using namespace std;
using namespace __gnu_pbds;

template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

ordered_set<ll> ors;

int dx[4]={1,0,-1,0};
int dy[4]={0,1,0,-1};
ld PI=3.14159265359;

ll power(ll a, ll n, ll m)
{
    ll ans=1;
    while(n)
    {
        if(n&1) ans=(ans*a)%m;
        n>>=1;
        a=(a*a)%m;
    }
    return ans;
}

ll invmod(ll a, ll m)
{
    return power(a,m-2,m);
}

void solve(ll tt) {
    ll n;
    cin>>n;
    ll a[n-1],b[n-1];
    repf(i,0,n-1) cin>>a[i];
    repf(i,0,n-1) cin>>b[i];
    ll t1=-1,t2=-1;
    repf(i,0,4) repf(j,0,4) {
        if((i|j)==a[0] && (i&j)==b[0]) {
            t1=i,t2=j;
            break;
        }
    } 
    if(t1==-1) {
        cout<<"NO\n";
        return;
    }
    bool ok=true;
    vector<ll> t={t1,t2};
    repf(i,1,n-1) {
        ll val=-1;
        repf(j,0,4) {
            if((t[i]|j)==a[i] && (t[i]&j)==b[i]) {
                val=j;
                break;
            }
        }
        if(val==-1) {
            ok=false;
            break;
        }
        t.pb(val);
    }
    if(ok) {
        cout<<"YES\n";
        for(auto i:t) cout<<i<<" ";
        cout<<endl;
        return;
    }
    t.clear();
    t.pb(t2);t.pb(t1);
    ok=true;
    repf(i,1,n-1) {
        ll val=-1;
        repf(j,0,4) {
            if((t[i]|j)==a[i] && (t[i]&j)==b[i]) {
                val=j;
                break;
            }
        }
        if(val==-1) {
            ok=false;
            break;
        }
        t.pb(val);
    }
    if(ok) {
        cout<<"YES\n";
        for(auto i:t) cout<<i<<" ";
        cout<<endl;
        return;
    }
    cout<<"NO\n";
} 


int main()
{
    IOS;
    ll t=1;
    // cin>>t;
    repf(tt,1,t+1) {
        solve(tt);
    }
}     