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 <image>.

Expression <image> 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
import java.util.*;
import java.io.*;
import java.math.*;
public class Solution{
    
    static int largest_power(int n){
        n = n|(n>>1);
        n = n|(n>>2);
        n = n|(n>>4);
        n = n|(n>>8);
        n = n|(n>>16);
        return (n+1)>>1;
    }
    
    static boolean check(int n,int i){
        if((n&(1<<i))!=0) return true;
        else return false;
    }
    
    static int up(int n){
        int two = largest_power(n);
        two *= 2;
        two--;
        return two;
    }
    
    public static void main(String[] args)throws IOException{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine());
        
        PrintWriter out = new PrintWriter(System.out);
        
        int n = Integer.parseInt(st.nextToken());
        
        
        int[] p = new int[n+1];
        for(int i=n;i>0;i--){
            if(p[i]>i) continue;
            p[i] = up(i) - i;
            if(p[p[i]]!=0 || (i&p[i])!=0) System.out.println("Damm neger!");
            else p[p[i]] = i;
        }
        long sum = 0;
        for(int i=0;i<=n;i++){
            sum += (long) (i^p[i]);
        }
        out.println(sum);
        for(int i=0;i<=n;i++){
            out.print(p[i]+" ");
        }
        out.println("");
        
        
        out.flush();
    }
    
}