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;
void Solve();
int main() { Solve(); }
int n, a[1000100], ans, p[1000];
void fuckzoe() {
  cin >> n;
  for (int i = 0; i <= n; ++i) a[i] = i;
  do {
    int t = 0;
    for (int i = 0; i <= n; ++i) t += (a[i] ^ i);
    if (t > ans) {
      ans = t;
      for (int j = 0; j <= n; ++j) p[j] = a[j];
    }
  } while (next_permutation(a, a + n + 1));
  cout << ans << endl;
  for (int i = 0; i <= n; ++i) cout << p[i] << ' ';
  cout << endl;
}
void Solve() {
  cin >> n;
  int C = 0;
  for (int i = n; i; i >>= 1, ++C)
    ;
  for (int i = n; i; --i) {
    if (a[i]) continue;
    int b[50] = {}, t = 0;
    for (int j = i, q = 0; j; t += ((j % 2) ^ 1) * (1 << q), j >>= 1, q++)
      ;
    a[i] = t;
    a[t] = i;
  }
  for (int i = n; i; --i)
    if (!a[i]) a[0] = i;
  long long ans = 0;
  for (int i = 0; i <= n; ++i) ans += (a[i] ^ i);
  cout << ans << endl;
  for (int i = 0; i <= n; ++i) cout << a[i] << ' ';
}
