題干：

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?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

題目大意：

? ?給n個(gè)數(shù)，分成兩組A和B，A中的可以找B中的連線，求最優(yōu)匹配，權(quán)值為二者的異或值。（

解題報(bào)告：

? 我把題目大意搞得跟二分圖一樣，，，但是其實(shí)不是的。。只是方便理解（看數(shù)據(jù)量也知道不是）

? 可以證明，一定對(duì)于每一個(gè)數(shù)都找到一個(gè)對(duì)應(yīng)的匹配

Since we need to maximize the result, we need to find such permutation, for which the least number of bit disappear. (We consider bit disappeared if it was 1 both in?i?and?pi, so in??it is 0). It turns out that for each?n?there is such permutation that no bit disappear. How to build it? We will be solving problem by iterations while?n?>?0. On each iteration, we need to find the biggest (the leftmost in binary representation) bit which is not?0?in binary representation of?n?and denote it position (bits are numbered from 0) by?b. Now we need to find integer?m?— minimal integer from?0?to?n, inclusive, such that?b-th bit is also?1?in it. After that you can see (look image below), that at??no bit disappear, at??no bit disappear, ..., at??no bit disappear. So, it is good to assign exactly that integers to our permutation, i. e.?pm?=?m?-?1?and?pm?-?1?=?m,?pm?+?1?=?m?-?2?and?pm?-?2?=?m?+?1?and so on. After that assign value?m?-?(n?-?m?+?1)?-?1?to?n?and go to next iteration.

Now when we know how to build permutation and that no bit disappear, the value of the answer is equal to?.

AC代碼：

#include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #define ll long long using namespace std; const int MAX=1e6 + 5; int ans[MAX]; ll sum=0,x=1; int main() {int n;cin >>n;while(x<=n) x<<=1;x--; for(int i=n; i>=0; i--) {if(ans[i]!=0) continue;while((x^i)>n||ans[x^i]!=0) x>>=1;ans[x^i]=i;ans[i]=x^i;}sum=0;for(int i=0; i<=n; i++)sum+=i^ans[i];cout <<sum<<endl;for(int i=0; i<=n; i++) {printf("%d%c",ans[i],i == n ? '\n' : ' ');}return 0; }

?

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