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
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。
题解：从大往小 找每个数二进制位数相同并全去都是1的数 与该数的异或值 就是该数的 顺次 ，然后标记异或得到的数
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
using namespace std;
const int N=1e6+10;
typedef long long ll;
int n;
int a[N];
int main()
{
scanf("%d",&n);
ll ans=0;
for(int i=n;i>=1;i--)
{
if(a[i]) continue;
int m=i,pos=0;
while(m)
{
m=m/2;
pos++;
}
int cnt=1<<pos;
cnt--;
a[i]=cnt-i;
a[cnt-i]=i;
ans+=cnt*2;
}
if(!a[0]) a[0]=0;
cout<<ans<<endl;
for(int i=0;i<=n;i++)printf("%d%c",a[i]," \n"[i==n]);
return 0;
}