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.
/*
         --┬--  |     |  --┬--                |       |
           |    |\    |    |                  |       |
           |    | \   |    |     ----->       |       |
           |    |  \  |    |                  |       |
           |    |   \ |    |                  |       |
         --┴--  |    \|    |                  └----   └----

*/

// #define pragma

#ifdef pragma
#pragma GCC optimize("Ofast")
#pragma GCC optimize("no-stack-protector")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
#pragma GCC diagnostic ignored "-Wunused-result"
#endif // pragma

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

#define ll long long
#define all(x) begin(x), end(x)
#define pb push_back
#define x first
#define y second
#define int long long
#define zero(x) memset(x, 0, sizeof(x))

using namespace std;
using namespace __gnu_pbds;


typedef vector<int> vi;
typedef vector<bool> vb;
typedef pair<int, int> pii;
typedef long double ld;
typedef vector<vi> matrix;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;

const int INF = 1e9 + 7;
const int MAXN = 1e6 + 7;
const ld EPS = 1e-6;
const ld PI = atan2(0, -1);

void seriy() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    cout << fixed << setprecision(10);
    #ifdef _android
        freopen("input", "r", stdin);
        freopen("output", "w", stdout);
    #endif
}

vi ans;
bool f = 0;
vi a(MAXN), b(MAXN);
int n;

void rec(int it) {
    if(it == n) {
        f = 1;
        return;
    }
//    cout << it << "     ";
//    for(auto i : ans) {
//        cout << i << " ";
//    }
//    cout << endl;
    int k = ans.back();
    for(int i = 0; i < 4; i++) {
        if((k | i) == a[it] && (k & i) == b[it]) {
            ans.pb(i);
            rec(it + 1);
            if(!f) ans.pop_back();
        }
    }
}

signed main() {
    seriy();

    cin >> n;

    for(int i = 0; i < n - 1; i++) {
        cin >> a[i];
    }
    for(int j = 0; j < n - 1; j++) {
        cin >> b[j];
    }
    for(int i = 0; i < 4; i++) {
        ans.pb(i);
        rec(0);
        if(f) {
            cout << "YES\n";
            ans.pop_back();
            for(auto i : ans) {
                cout << i << " ";
            }
            return 0;
        }
        else {
            ans.pop_back();
        }
    }
    cout << "NO";
    return 0;
}
