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 a[1000005], pointer = 1;
long long ans;
int tavan2(int x) {
  int tmp = 1;
  for (int i = 0; i < x; i++) tmp *= 2;
  return tmp;
}
void tabe(int x) {
  bool d = 1;
  int tmp = 1;
  if (x == -1) return;
  if (x == 0) {
    a[pointer] = 0;
    pointer++;
    return;
  }
  if (x == 1) {
    a[pointer] = 0;
    a[pointer + 1] = 1;
    ans += 2;
    pointer += 2;
    return;
  }
  for (int i = 1; d == 1; i++) {
    tmp = i;
    if (tavan2(i) > x) d = 0;
  }
  tmp--;
  tmp = tavan2(tmp);
  for (int i = x; i >= ((2 * tmp) - x - 1); i--) {
    ans += ((2 * tmp) - 1);
    a[pointer] = (((2 * tmp) - 1) - i);
    pointer++;
  }
  tabe(x - (2 * (x - tmp + 1)));
}
int main() {
  int n;
  cin >> n;
  tabe(n);
  cout << ans << endl;
  for (int i = pointer - 1; i > 0; i--) cout << a[i] << " ";
  cout << endl;
  return 0;
}
