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 resp[1000010];
int v[1000010];
bool usado[1000010];
int n;
int nbits(int n) {
  int r = 0;
  while (n) {
    r++;
    n >>= 1;
  }
  return r;
}
int main() {
  scanf("%d", &n);
  n++;
  memset(v, 0xff, sizeof(v));
  int nb = nbits(n - 1);
  int mask = (1 << nb) - 1;
  int nusa = nb;
  int tem = 0;
  memset(usado, false, sizeof(usado));
  while (tem < n) {
    for (int i = n - 1; i >= 0; i--)
      if (v[i] == -1) {
        int inv = (i ^ mask) & ((1 << nusa) - 1);
        if (inv < n and !usado[inv]) {
          usado[inv] = true;
          v[i] = inv;
          tem++;
        }
      }
    nusa--;
  }
  long long int soma = 0;
  for (int i = 0; i < n; i++) soma = soma + (long long int)(v[i] ^ i);
  printf("%I64d\n", soma);
  for (int i = 0; i < n; i++) printf("%d ", v[i]);
  printf("\n");
  return 0;
}
