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 N = 1000010;
const int L = 25;
int n;
int f[L + 1], ans[N];
long long tans = 0;
void init() {
  f[0] = 1;
  for (int i = 1; i <= L; i++) f[i] = f[i - 1] * 2;
}
int get(int m) {
  int res = L;
  while (f[res] > m) res--;
  return res;
}
int main() {
  int i;
  init();
  scanf("%d", &n);
  int tmp = n;
  while (tmp > 0) {
    int t = get(tmp);
    for (i = 0; f[t] + i <= tmp; i++) {
      ans[f[t] + i] = f[t] - i - 1;
      ans[f[t] - i - 1] = f[t] + i;
    }
    tmp -= (tmp - f[t] + 1) * 2;
  }
  for (i = 0; i <= n; i++) tans += ans[i] ^ i;
  printf("%I64d\n", tans);
  for (i = 0; i <= n; i++) printf("%d ", ans[i]);
  return 0;
}
