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;
vector<int> a;
void build(int start, int k) {
  if (k == 0) return;
  build(start, k - 1);
  build(start + (1 << (k - 1)), k - 1);
  for (int i = 0; i < (int)(1 << (k - 1)); i++) {
    swap(a[start + i], a[start + (1 << (k - 1)) + i]);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  n++;
  a.resize(n);
  iota(a.begin(), a.end(), 0);
  int start = 0;
  while (start < n) {
    int k;
    for (k = 0; start + (1 << (k + 1)) <= n; k++)
      ;
    build(start, k);
    vector<int> where(1 << k);
    for (int i = start; i < start + (1 << k); i++) {
      where[a[i]] = i;
    }
    for (int i = start + (1 << k); i < n; i++) {
      swap(a[i], a[where[a[i] - (1 << k)]]);
    }
    start += (1 << k);
  }
  long long ans = 0;
  for (int i = 0; i < (int)(n); i++) {
    ans += i ^ a[i];
  }
  cout << ans << endl;
  for (int i = 0; i < (int)(n); i++) cout << a[i] << " ";
  cout << endl;
  return 0;
}
