Number Sequence HDU - 5014

There is a special number sequence which has n+1 integers. For each number in sequence, we have two rules:

● a i ∈ 0,n0,n

● a i ≠ a j( i ≠ j )

For sequence a and sequence b, the integrating degree t is defined as follows(“�” denotes exclusive or):

t = (a 0 � b 0) + (a 1 � b 1) +・・・+ (a n � b n)

(sequence B should also satisfy the rules described above)

Now give you a number n and the sequence a. You should calculate the maximum integrating degree t and print the sequence b.

Input

There are multiple test cases. Please process till EOF.

For each case, the first line contains an integer n(1 ≤ n ≤ 10 5), The second line contains a 0,a 1,a 2,…,a n.

Output

For each case, output two lines.The first line contains the maximum integrating degree t. The second line contains n+1 integers b 0,b 1,b 2,…,b n. There is exactly one space between b i and b i+1 (0 ≤ i ≤ n - 1). Don’t ouput any spaces after b n.

Sample Input

4

2 0 1 4 3

Sample Output

20

1 0 2 3 4

#include<map>
#include<set>
#include<queue>
#include<stack>
#include<vector>
#include<math.h>
#include<cstdio>
#include<sstream>
#include<numeric>//STL数值算法头文件
#include<stdlib.h>
#include <ctype.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<functional>//模板类头文件
using namespace std;

typedef long long ll;
const int maxn=110000;
const int INF=0x3f3f3f3f;

ll a[maxn];
ll d[maxn];
int main()
{
    ll n;
    while(~scanf("%I64d",&n))
    {
        for(ll i=0; i<=n; i++)
            scanf("%I64d",&a[i]);
        memset(d,-1,sizeof(d));
        ll ans=0;
        for(ll i=n; i>=0; i--)
        {
            ll t=0;
            if(d[i]==-1)
            {
                for(ll j=0;; j++)
                {
                    if(!(i&(1<<j)))  t+=(1<<j);
                    if(t>=i)
                    {
                        t-=(1<<j);
                        break;
                    }
                }
                ans+=(i^t)*2;
                d[i]=t;
                d[t]=i;
            }
        }
        printf("%I64d\n",ans);
        for(ll i=0; i<=n; i++)
            printf(i==n?"%I64d\n":"%I64d ",d[a[i]]);
    }
    return 0;
}