Polo the Penguin and XOR operation(位运算 思维)

Polo the Penguin and XOR operation

 CodeForces - 288C 

Little penguin Polo likes permutations. But most of all he likes permutations of integers from 0 to n, inclusive.

For permutation p = p₀, p₁, ..., p_n, Polo has defined its beauty — number .

Expression  means applying the operation of bitwise excluding "OR" to numbers xand y. This operation exists in all modern programming languages, for example, in language C++ and Java it is represented as "^" and in Pascal — as "xor".

Help him find among all permutations of integers from 0 to n the permutation with the maximum beauty.

Input

The single line contains a positive integer n (1 ≤ n ≤ 10⁶).

Output

In the first line print integer m the maximum possible beauty. In the second line print any permutation of integers from 0 to n with the beauty equal to m.

If there are several suitable permutations, you are allowed to print any of them.

Examples

Input

4

Output

20
0 2 1 4 3
题意：给定一个 n (1 <= n <= 10^6)，求(0 ^ p0) + (1 ^ p1) + (2 ^ p2) +…… + (n ^ pn) 的最大值，其中p0 ~ pn为 0 ~ n 中的数，且每个数只利用一次。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int maxn=1e6+10;
int n,ans[maxn];
int main()
{
    scanf("%d",&n);
    memset(ans,-1,sizeof(ans));
    for(int i=n;i>=0;i--)
    {
        if(ans[i]!=-1)
            continue;
        int k1=log2(i)+1;//求出每个数的二进制位数 并加一位 
        int k2=(1<<k1)-1;//找到和二进制下全为1的数且位数相同。
        //就比如说你知道了二进制下位数为三且全为1的数 111 
        //然后你求的其实就是两个位数为三的数异或为111，而且你知道了其中的一个
        //例如6 二进制位110 就是在求110^谁=111 那么这个谁 就相当于111^110  
        
        ans[k2^i]=i;
        ans[i]=k2^i;
    }
    long long sum=0;
    for(int i=0;i<=n;i++)
    {
        sum+=(long long )(i^ans[i]); 
    }
    printf("%lld\n",sum);
    for(int i=0;i<=n;i++)
    {
        if(i)
        {
            printf(" ");
        }
        printf("%d",ans[i]);
    }
    printf("\n");
}