題目：

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

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

解題報告：思維題目，之前一開始就走錯了方向，自己找錯了結論，以為就是討論n的奇偶性，若是n為奇數的話，就可以和0異或，之間的相鄰的兩位奇偶進行異或，若n為偶數的話，直接排除0的干擾，從1-n之間奇偶進行異或，確實是在小範圍的測試樣例是可以正確實現的。其實最終的正確的解法是：先找到最大的全一數字，然後尋找合適的進行，最大異或，然後逐次尋找進行最優判斷，直到全部進行完畢。

ac程式碼：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
#define mod 1000000007

const int maxn =1e7+100;
ll num[maxn];
int main()
{
	ll n;
	while(scanf("%lld",&n)!=EOF)
	{
		ll t=1;
		while(t<=n)
			t*=2;
		t--;
		memset(num,-1,sizeof(num));
		for(int i=n;i>=0;i--)
		{
			if(num[i]!=-1)
				continue;
			else
			{
				while((t^i)>n||num[t^i]!=-1)
					t/=2;//出界
				num[i]=t^i;
				num[t^i]=i;//交換，最優交換。
			}
		}
		ll sum=0;
		for(int i=0;i<=n;i++)
			sum+=(i^num[i]);
		printf("%lld\n",sum);
		for(int i=0;i<n;i++)
			printf("%lld ",num[i]);
		printf("%d\n",num[n]);
	}
	return 0;
}