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;

}