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
#include <bits/stdc++.h>
using namespace std;
int ans[1000005];
int two[100];
int gettwo(int n) {
  int i;
  for (i = 0; i < n; i++) {
    if (two[i] <= n && two[i + 1] > n) return two[i + 1] - 1;
  }
}
int main() {
  int i, n, k, gt;
  long long sum = 0;
  two[0] = 1;
  for (i = 1; i < 26; i++) two[i] = 2 * two[i - 1];
  scanf("%d", &n);
  if (n == 1) {
    printf("2\n1 0\n");
    return 0;
  }
  for (i = n; i > 0; i--) {
    if (ans[i] == 0) {
      gt = gettwo(i);
      sum += 2 * gt;
      ans[i] = (i ^ gt);
      ans[(i ^ gt)] = i;
    } else
      continue;
  }
  printf("%I64d\n", sum);
  for (i = 0; i <= n; i++) {
    if (i != 0) printf(" ");
    printf("%d", ans[i]);
  }
  return 0;
}
