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 int MaxN = (int)1e6 + 10;
int a[21], b[MaxN], n;
bool used[MaxN];
int poisk(int n) {
  int kol = 0;
  while (n) {
    n /= 2;
    kol++;
  }
  return kol;
}
void print() {
  for (int i = 0; i <= n; i++) {
    cout << b[i] << " ";
  }
  cout << endl;
}
int main() {
  cin >> n;
  a[0] = 1;
  for (int i = 0; i <= n; i++) {
    b[i] = 0;
    used[i] = 0;
  }
  for (int i = 1; i <= 20; i++) {
    a[i] = a[i - 1] * 2;
  }
  for (int i = n; i >= 0; i--) {
    if (used[i] == 0) {
      int ind = poisk(i);
      int tmp = a[ind] - 1 - i;
      b[i] = tmp;
      b[tmp] = i;
      used[tmp] = 1;
      used[i] = 1;
    }
  }
  long long ans = 0;
  for (int i = 0; i <= n; i++) {
    ans = ans + (long long)(b[i] | i);
  }
  printf("%I64d\n", ans);
  for (int i = 0; i <= n; i++) {
    printf("%d ", b[i]);
  }
  cin.get();
  cin.get();
}
