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 vis[1010100];
int num[1010101];
int main() {
  int n, k;
  long long sum = 0;
  scanf("%d", &n);
  int one[100] = {0};
  one[1] = 1;
  for (int i = 2; i <= 24; i++) {
    one[i] = (one[i - 1] << 1) + 1;
  }
  int tt = *lower_bound(one, one + 23, n);
  for (int st = tt - n, ed = n; st < ed; ed--, st++) {
    sum += (st ^ ed) * 2;
    num[st] = ed;
    num[ed] = st;
    vis[st] = 1;
    vis[ed] = 1;
  }
  int ne = tt - n - 1;
  while (1) {
    int tt = *lower_bound(one, one + 23, ne);
    int ff = tt - ne;
    for (int st = tt - ne, ed = ne; st < ed; ed--, st++) {
      sum += (st ^ ed) * 2;
      num[st] = ed;
      num[ed] = st;
      vis[st] = 1;
      vis[ed] = 1;
    }
    ne = tt - ne - 1;
    if (ff == 0 || ne < 0) break;
  }
  printf("%lld\n", sum);
  printf("%d", num[0]);
  for (int i = 1; i <= n; i++) {
    printf(" %d", num[i]);
  }
  putchar('\n');
}
