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;
template <typename T>
T gcd(T a, T b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
template <typename T>
T pow(T a, T b, long long m) {
  T ans = 1;
  while (b > 0) {
    if (b % 2 == 1) ans = ((ans % m) * (a % m)) % m;
    b /= 2;
    a = ((a % m) * (a % m)) % m;
  }
  return ans % m;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int n;
  cin >> n;
  vector<int> v(n + 1, -1);
  long long ans = 0;
  for (int i = n; i >= 0; i--) {
    if (v[i] != -1) continue;
    bool fnd = false;
    long long temp = 0;
    long long temp1 = 0;
    for (int j = 21; j >= 0; j--) {
      if (i & (1LL << j)) {
        fnd = 1;
        temp += (1LL << j);
      } else {
        if (fnd) {
          temp += (1LL << j);
          temp1 += (1LL << j);
        }
      }
    }
    ans += 2 * temp;
    v[i] = temp1;
    v[temp1] = i;
  }
  cout << ans << "\n";
  for (int i = 0; i <= n; i++) cout << v[i] << " ";
}
