CF 177（div2） E（greedy ）

E. 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, ..., p_n, 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.

Sample test(s)

Input

4

Output

 

因为亦或就是不进位的二进制加法，所以很容易想到尽量不要让每个位上的1浪费掉，就有了一个贪心的想法，从大到小，用全1的数去减当前的这个数，就是这个数应该对应的数，结合样例测试后的确是可行的。但是发现，对较小的数这样做，得到的对应数可能大于最大值，没法对每个数都用一个统一的方案解决，怎么办呢？其实可以发现，较小的数对应的数，在对称的那个较大数求较小数时应经就找到了，就是说我们不用一次次去找每个数对应的数，而是每次找一对数，这样问题就解决了。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <stack>
#define LL long long
using namespace std;
const int maxn = 1000000 + 5;

LL pre[32];
int ans[maxn];

LL len(int x){
    LL ret = 0;
    while(x){
        ret++;
        x = x/2;;
    }
    return ret;
}

int main(){
    int n;
    LL total;
    pre[0] = 0;
    pre[1] = 1;
    for(int i = 2;i < 32;i++) pre[i] = (pre[i-1]<<1) + 1;
    while(cin >> n){
        memset(ans,-1,sizeof(ans));
        total = 0;
        for(int i = n;i >= 0;i--){
            if(ans[i] == -1){
                int j = pre[len(i)]-i;
                ans[i] = j;
                ans[j] = i;
                total += 2*pre[len(i)];
            }
        }
        cout << total << endl;
        for(int i = 0;i <= n;i++) printf("%d ",ans[i]);
        cout << endl;
    }
    return 0;
}