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 table[1 << 20] = {0};
int answer[1 << 20] = {0};
int main() {
  ios_base::sync_with_stdio(false);
  long long n;
  cin >> n;
  int m = ++n;
  for (int i = 0; i < n; ++i) {
    table[i] = i;
  }
  int *ptr = answer;
  while (m != 0) {
    int msb = 0;
    while ((m >> (msb + 1)) > 0) {
      ++msb;
    }
    int k = m - (1 << msb);
    for (int i = 0; i < (1 << msb) - k; ++i) {
      ptr[i] = ((1 << msb) - 1) & ~i;
    }
    for (int i = (1 << msb) - k; i < (1 << msb); ++i) {
      ptr[i] = (1 << msb) | (((1 << msb) - 1) & ~i);
    }
    m ^= (1 << msb);
    ptr += (1 << msb);
  }
  long long sum = 0;
  for (int i = 0; i < n; ++i) {
    sum += i ^ answer[i];
  }
  cout << sum << endl;
  for (int i = 0; i < n; ++i) {
    if (i != 0) {
      cout << " ";
    }
    cout << answer[i];
  }
  cout << endl;
  return 0;
}
