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;
const long long MAXN = (1LL << 62);
const int INF = (1 << 30);
const double PI = acos(-1.0);
const double EPS = 1e-11;
int twoPow[21];
int n;
int perm[1000005];
int solve() {
  int i, ii;
  for (i = 0; i <= 20; i++) twoPow[i] = (1 << i);
  i = n;
  while (i > 0) {
    int k;
    for (ii = 0; ii <= 20; ii++)
      if (i < twoPow[ii]) break;
    k = twoPow[ii - 1];
    for (ii = k; ii <= i; ii++) {
      perm[ii] = 2 * k - 1 - ii;
      perm[2 * k - 1 - ii] = ii;
    }
    i = 2 * k - 2 - i;
  }
  if (i == 0) perm[0] = 0;
  return 0;
}
int main() {
  long long res;
  scanf("%d", &n);
  res = n;
  res *= (n + 1);
  printf("%I64d\n", res);
  solve();
  for (int i = 0; i <= n; i++) printf("%d ", perm[i]);
  printf("\n");
  return 0;
}
