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;
long long rdtsc() {
  long long tmp;
  asm("rdtsc" : "=A"(tmp));
  return tmp;
}
const int maxn = (int)1e6 + 1;
int a[maxn];
long long solve(int n) {
  if (n < 0) return 0;
  if (!n) {
    a[0] = 0;
    return 0;
  }
  int k = 0;
  for (; (1 << (k + 1)) <= n; k++)
    ;
  int it1, it2;
  long long ans = 0;
  for (it1 = (1 << k) - 1, it2 = (1 << k); it1 >= 0 && it2 <= n; it1--, it2++) {
    a[it1] = it2, a[it2] = it1;
    ans += 2 * (it1 ^ it2);
  }
  return ans + solve(it1);
}
int main() {
  srand(rdtsc());
  int n;
  while (scanf("%d", &n) >= 1) {
    printf(
        "%lld"
        "\n",
        solve(n));
    for (int i = 0; i <= n; i++) printf("%d%c", a[i], " \n"[i == n]);
  }
  return 0;
}
