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
// practice with kaiboy
import java.io.*;
import java.util.*;

public class CF289E extends PrintWriter {
	CF289E() { super(System.out, true); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF289E o = new CF289E(); o.main(); o.flush();
	}

	void reverse(int[] aa, int l, int r) {
		for (int i = l, j = r - 1; i < j; i++, j--) {
			int tmp = aa[i]; aa[i] = aa[j]; aa[j] = tmp;
		}
	}
	void main() {
		int n = sc.nextInt();
		long ans = 0;
		int[] aa = new int[n + 1];
		for (int i = 0; i <= n; i++)
			aa[i] = i;
		int n_ = n;
		for (int h = 20; n > 0; ) {
			while (h >= 0 && (n & 1 << h) == 0)
				h--;
			int m = n - ((1 << h) - 1);
			reverse(aa, n - m * 2 + 1, n + 1);
			n -= m * 2;
			ans += m * 2L * ((1 << h + 1) - 1);
		}
		println(ans);
		for (int i = 0; i <= n_; i++)
			print(aa[i] + " ");
		println();
	}
}
