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;
template <class stl>
void DBGSTL(stl a) {
  for (__typeof((a).begin()) i = (a).begin(); i != (a).end(); i++) {
    cerr << *i << " ";
  }
  cerr << "\n";
  return;
}
vector<int> power;
int p[1000005];
int main() {
  int n, x;
  while (cin >> n) {
    ;
    x = n;
    x++;
    memset((p), (-1), sizeof(p));
    long long res = 0;
    if (n == 0) {
      cout << 0 << endl << 0 << endl;
      return 0;
    }
    if (!(x & (x - 1)) && x) {
      for (int i = n; i >= 0; i--) {
        if (p[i] != -1) {
          continue;
        }
        if (i == n) {
          res += 2 * n;
          p[n] = 0;
          p[0] = n;
        } else {
          int v = i;
          int high = 0;
          for (int j = 0; j <= 20; j++) {
            if (v & (1 << j)) {
              high = j;
            }
          }
          for (int j = 0; j <= high; j++) {
            v ^= (1 << j);
          }
          int val = (1 << high) - 1;
          res += (2 * (v ^ i));
          p[i] = v;
          p[v] = i;
        }
      }
    } else {
      for (int i = n; i >= 0; i--) {
        if (p[i] != -1) {
          continue;
        }
        if (i == 0) {
          p[i] = 0;
          continue;
        }
        int v = i;
        int high = 0;
        for (int j = 0; j <= 20; j++) {
          if (v & (1 << j)) {
            high = j;
          }
        }
        for (int j = 0; j <= high; j++) {
          v ^= (1 << j);
        }
        high++;
        int val = (1 << high) - 1;
        res += (2 * (v ^ i));
        p[i] = v;
        p[v] = i;
      }
    }
    cout << res << "\n";
    for (int i = 0; i <= n; i++) {
      if (i) printf(" ");
      printf("%d", p[i]);
    }
    printf("\n");
  }
  return 0;
}
