CodeForce 517 Div 2. B Curiosity Has No Limits

http://codeforces.com/contest/1072/problem/B

B. Curiosity Has No Limits

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

When Masha came to math classes today, she saw two integer sequences of length n−1n−1 on the blackboard. Let's denote the elements of the first sequence as aiai (0≤ai≤30≤ai≤3), and the elements of the second sequence as bibi (0≤bi≤30≤bi≤3).

Masha became interested if or not there is an integer sequence of length nn, which elements we will denote as titi (0≤ti≤30≤ti≤3), so that for every ii (1≤i≤n−11≤i≤n−1) the following is true:

• ai=ti|ti+1ai=ti|ti+1 (where || denotes the bitwise OR operation) and
• bi=ti&ti+1bi=ti&ti+1 (where && denotes the bitwise AND operation).

The question appeared to be too difficult for Masha, so now she asked you to check whether such a sequence titi of length nn 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 nn (2≤n≤1052≤n≤105) — the length of the sequence titi.

The second line contains n−1n−1 integers a1,a2,…,an−1a1,a2,…,an−1 (0≤ai≤30≤ai≤3) — the first sequence on the blackboard.

The third line contains n−1n−1 integers b1,b2,…,bn−1b1,b2,…,bn−1 (0≤bi≤30≤bi≤3) — the second sequence on the blackboard.

Output

In the first line print "YES" (without quotes), if there is a sequence titi 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 nn integers t1,t2,…,tnt1,t2,…,tn (0≤ti≤30≤ti≤3) — the sequence that satisfies the statements conditions.

If there are multiple answers, print any of them.

Examples

input

Copy

4
3 3 2
1 2 0

output

Copy

YES
1 3 2 0 

input

Copy

3
1 3
3 2

output

Copy

NO

Note

In the first example it's easy to see that the sequence from output satisfies the given conditions:

• t1|t2=(012)|(112)=(112)=3=a1t1|t2=(012)|(112)=(112)=3=a1 and t1&t2=(012)&(112)=(012)=1=b1t1&t2=(012)&(112)=(012)=1=b1;
• t2|t3=(112)|(102)=(112)=3=a2t2|t3=(112)|(102)=(112)=3=a2 and t2&t3=(112)&(102)=(102)=2=b2t2&t3=(112)&(102)=(102)=2=b2;
• t3|t4=(102)|(002)=(102)=2=a3t3|t4=(102)|(002)=(102)=2=a3 and t3&t4=(102)&(002)=(002)=0=b3t3&t4=(102)&(002)=(002)=0=b3.

In the second example there is no such sequence.

题意：求一个n长度的t数组，满足t[i]&t[i+1] == b[i]   同时 t[i]|t[i+1] == a[i]。 a,b数组长度为n-1

枚举法，想过枚举但是没有想到是这样做的。因为t数组中，最后一个是最特殊的，只与a和b数组中的一个元素相关（自己没有发现）。同时没有发现如果确定了一个点，那么下一个点的值是唯一确定的。这样想来，其实枚举第一个也是可行的了。

同时，需要记住，如果t[i]确定了，那么与不同的t[i+1]进行或，与运算一定会得到的是不同的结果。

和之前做过的一道Fliptile有些像。

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <stack>
#define ll long long
//#define local

using namespace std;

const int MOD = 1e9+;
const int inf = 0x3f3f3f3f;
const double PI = acos(-1.0);
const int maxn = (1e5+);
const int maxedge = *;
int a[maxn], b[maxn];
int t[maxn];

int main() {
#ifdef local
    if(freopen("/Users/Andrew/Desktop/data.txt", "r", stdin) == NULL) printf("can't open this file!\n");
#endif
    int n;
    scanf("%d", &n);
    for (int i = ; i < n-; ++i) {
        scanf("%d", a+i);
    }
    for (int i = ;i < n-; ++i) {
        scanf("%d", b+i);
    }
    for (int i = ; i < ; ++i) {
        memset(t, -, sizeof(t));
        t[n-] = i;
        for (int j = n-; j >= ; --j) {
            for (int k = ; k < ; ++k) {
                if ((k|t[j+])==a[j] && (k&t[j+])==b[j]) {
                    t[j] = k;
                    break;
                }
            }
            if (t[j] == -) break;
        }
        if (t[] != -) break;
    }
    if (t[] == -) printf("NO\n");
    else {
        printf("YES\n");
        for (int i = ; i < n; ++i) {
            printf("%d", t[i]);
            if (i != n-) printf(" ");
        }
        printf("\n");
    }
#ifdef local
    fclose(stdin);
#endif
    return ;
}