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;
inline bool insize(int c, int l, int r) {
  if (c >= l && c <= r) return true;
  return false;
}
template <class T>
inline void checkmin(T &a, T b) {
  if (a == -1 || a > b) a = b;
}
template <class T>
inline void checkmax(T &a, T b) {
  if (a < b) a = b;
}
int dx[] = {-1, 0, 1, 0};
int dy[] = {0, 1, 0, -1};
const int N = 1000005;
const int E = 200555;
const int INF = 0x3f3f3f3f;
const long long LINF = 0x3F3F3F3F3F3F3F3FLL;
int x[N], y[N];
vector<int> vi;
void dfs(int n) {
  if (n < 0) return;
  int t = 1;
  for (; t <= n; t *= 2)
    ;
  for (int i = t / 2; i <= n; i++) {
    y[i] = t - i - 1;
    y[t - i - 1] = i;
  }
  dfs(t - n - 2);
}
int main() {
  int n;
  long long sum = 0;
  cin >> n;
  for (int(i) = (1); (i) <= (n); (i)++) {
    sum += 2 * i;
  }
  cout << sum << endl;
  dfs(n);
  for (int(i) = (0); (i) <= (n); (i)++) printf("%d ", y[i]);
  printf("\n");
  ;
}
