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 conv(int n) {
  int ans = 0;
  vector<int> bits;
  while (n != 0) {
    if (n & 1)
      bits.push_back(0);
    else
      bits.push_back(1);
    n = n >> 1;
  }
  for (int i = bits.size() - 1; i >= 0; i--) ans = (ans * 2) + bits[i];
  return ans;
}
int main() {
  int n;
  while (scanf("%d", &n) != EOF) {
    int counter = n;
    vector<int> mark(n + 1, -1);
    while (n >= 0) {
      if (mark[n] == -1) {
        int solve = conv(n);
        mark[n] = solve;
        mark[solve] = n;
      }
      n--;
    }
    long long ans = 0;
    for (int i = 0; i <= counter; i++) {
      ans += (i ^ mark[i]);
    }
    cout << ans << endl;
    for (int i = 0; i <= counter; i++) printf("%d ", mark[i]);
    printf("\n");
  }
  return 0;
}
