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;
int main() {
  int n;
  scanf("%d", &n);
  int p = 0;
  while ((1 << p) < n) p++;
  p = (1 << p);
  int xn = n;
  vector<int> an;
  an.clear();
  while (xn >= 0) {
    if (p > xn) {
      p >>= 1;
      continue;
    }
    if (p == 0) {
      an.push_back(0);
      xn--;
    } else {
      an.push_back(xn - (p - 1));
      xn -= an[an.size() - 1] * 2;
      p >>= 1;
    }
  }
  printf("%I64d\n", ((long long)n * (long long)(n + 1)));
  bool isNewNumber = true;
  int x = 0;
  int idx = an.size() - 1;
  while (x < n) {
    if (an[idx] == 0) {
      if (!isNewNumber) printf(" ");
      printf("%d", x);
      isNewNumber = false;
      x++;
    } else {
      for (int i = 0; i < an[idx] * 2; i++) {
        if (!isNewNumber) printf(" ");
        printf("%d", x + an[idx] * 2 - i - 1);
        isNewNumber = false;
      }
      x += an[idx] * 2;
    }
    idx--;
  }
  puts("");
  return 0;
}
