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.*;

public class E {
    public static Scanner in;
    public static PrintWriter out;
    public static long ans;

    public static void main(String args[]) {
        in = new Scanner(System.in);
        out = new PrintWriter(System.out);
        int n = in.nextInt();
        ans = 0;
        bt(n);
        out.println(ans);
        bt2(n);
        out.close();
    }

    public static void bt(int n) {
        if ((Math.log(n + 1) / Math.log(2)) % 1 == 0) {
            for (int i = n, j = 0; i >= 0; i--, j++) {
                ans += i ^ j; 
            }
            return;
        }
        int a = (int) Math.pow(2, (int) (Math.log(n) / Math.log(2)));
        int x = n - a + 1;
        int m = n - (2 * x);
        bt(m);
        for (int i = n, j = m + 1; i > m; j++, i--){
            ans += j ^ i;
        }
    }
    public static void bt2(int n) {
        if ((Math.log(n + 1) / Math.log(2)) % 1 == 0) {
            for (int i = n, j = 0; i >= 0; i--, j++) {
                out.print(i + " ");
            }
            return;
        }
        int a = (int) Math.pow(2, (int) (Math.log(n) / Math.log(2)));
        int x = n - a + 1;
        int m = n - (2 * x);
        bt2(m);
        for (int i = n, j = m + 1; i > m; j++, i--){
            out.print(i + " ");
        }
    }
}
