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 INF = 0x3F3F3F3F;
const int MAXN = 100001;
int arr[MAXN * 10];
bool mark[MAXN * 10];
pair<int, int> get_complement(int i) {
  int bits = 1;
  int k = 0;
  while (bits <= i) {
    bits *= 2;
    k++;
  }
  int s = (1 << k) - 1;
  int result = 0;
  for (int j = 0; j < k; ++j) {
    if (!(i & (1 << j))) {
      result += (1 << j);
    }
  }
  return make_pair(result, s);
}
inline void input() {
  int n;
  scanf("%d", &n);
  ;
  long long int result = 0;
  for (int i = n; i >= 0; --i) {
    if (mark[i]) continue;
    pair<int, int> p = get_complement(i);
    if (!mark[p.first]) {
      result += (2LL * (long long int)p.second);
      mark[p.first] = true;
      mark[i] = true;
      arr[i] = p.first;
      arr[p.first] = i;
    }
  }
  printf("%lld\n", result);
  for (int i = 0; i < n; ++i) {
    printf("%d ", arr[i]);
  }
  printf("%d\n", arr[n]);
}
int main() {
  input();
  return 0;
}
