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 pow(int a, int b) {
  int prod = 1;
  int i;
  for (i = 0; i < b; i++) {
    prod = prod * a;
  }
  return prod;
}
long long int j, ans[1000005], k;
void print(long long n) {
  while (n > 0) {
    j = 1;
    k = 0;
    while ((2 * j) <= n) j = j << 1;
    int curr = j;
    while ((curr + k) <= n) {
      ans[curr - k - 1] = curr + k;
      ans[curr + k] = curr - k - 1;
      k++;
    }
    n = curr - (n - curr + 1) - 1;
  }
}
int main() {
  long long i;
  long long n;
  cin >> n;
  long long n1 = n;
  long long l, rest = 1;
  long long sum = 0;
  while (1) {
    rest = 1;
    while ((2 * rest) <= n) {
      rest = rest << 1;
    }
    sum = sum + 2 * (n - rest + 1) * ((rest * 2) - 1);
    n = rest - (n - rest + 1) - 1;
    if (n <= 0) break;
  }
  n = n1;
  cout << sum << "\n";
  print(n);
  for (i = 0; i <= n; i++) cout << ans[i] << " ";
  return 0;
}
