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 double eps = 1e-15;
const int maxN = 1 * 100000 + 10, mod = (int)1e9 + 7;
int n;
string s;
vector<int> a;
void ms(int x, int y) {
  for (int i = 0; i < 2 * y; ++i) a.push_back((x - 2 * y + i));
}
pair<int, int> bo(int x) {
  int h = 1;
  while (h <= x) h *= 2;
  return make_pair(h / 2, x - h / 2);
}
int main(void) {
  ios_base::sync_with_stdio(0);
  cin >> n;
  cout << ((long long)n * (n + 1)) << endl;
  int k = n;
  pair<int, int> p;
  while (a.size() != n + 1) {
    p = bo(k);
    if (k != 0)
      ms(p.first + p.second + 1, p.second + 1);
    else
      a.push_back(0);
    k = p.first - p.second - 2;
  }
  for (int i = n + 1 - 1; i >= 0; --i) cout << a[i] << ' ';
  return 0;
}
