CodeForces 288C

C. Polo the Penguin and XOR operation

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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

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

Expression  means applying the operation of bitwise excluding "OR" to numbers x and 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

200 2 1 4 3

题意：求0~n的一排列，使得 .最大。

思路：贪心，2进制位数多的创造的价值最大，优先满足这些数字，10是1010，满足这个的最大的是0101也就是5，然后推下去就是9对应6,8对应7

然后4对应3，而2对应1，0给个0，就行了，注意数据大的时候会超INT。

#include<bits/stdc++.h>using namespace std;int vis[1024000];int main(){    int n;    scanf("%d",&n);    for(int i = n;i >= 1;i--)    {        if(vis[i])continue;        int lv = log2(i)+1;        int num = (1 << lv)-1;        int tmp = num ^ i;        vis[tmp] = i;        vis[i] = tmp;    }    long long sum = 0;    for(int i = 0;i <= n;i++)    {        sum+=i^vis[i];    }    printf("%lld\n",sum);    for(int i = 0;i <= n;i++)        printf("%d ",vis[i]);    return 0;}