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 dx[] = {-1, 0, 1, 0, -1, -1, 1, 1};
const int dy[] = {0, -1, 0, 1, -1, 1, 1, -1};
template <class T>
inline bool get(T& n) {
  char c, mul = 1;
  n = 0;
  while (!isdigit(c = getchar_unlocked()) && (c != EOF) && (c != '-'))
    ;
  if (c == EOF) return false;
  if (c == '-') {
    c = getchar_unlocked();
    mul = -1;
  }
  for (; isdigit(c); c = getchar_unlocked()) n = n * 10 + c - '0';
  n *= mul;
  return true;
}
inline void get(char& c) {
  while (isspace(c = getchar_unlocked()))
    ;
}
bitset<1000001> used;
int ans[1000001];
int main() {
  int n;
  get(n);
  long long sum = 0;
  int bit;
  for (bit = 0; 1 << bit <= n; ++bit)
    ;
  int full = (1 << bit) - 1;
  for (int(i) = (n), _ = (0); (i) >= _; --(i)) {
    int v = full ^ i;
    for (int(j) = (bit), _ = (0); (j) >= _; --(j))
      if ((v > n || used[v]) && (v & (1 << j))) v ^= (1 << j);
    used[v] = 1;
    ans[i] = v;
    sum += i ^ ans[i];
  }
  printf("%I64d\n", sum);
  for (int(i) = (0), _ = (n); (i) <= _; ++(i)) printf("%d ", ans[i]);
  return 0;
}
