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 MAXN = 1e6 + 100;
const int inf_int = 1e9 + 100;
const long long inf_ll = 1e18;
const double PI = 3.1415926535898;
bool debug = 0;
int T, K;
const int MOD = 1e9 + 7;
int ans[MAXN];
void solve() {
  int n;
  cin >> n;
  int n1 = n;
  while (n > 0) {
    int x = 1;
    while (x <= n) {
      x <<= 1;
    }
    x >>= 1;
    int mn = inf_int;
    int m = x;
    if (debug) cerr << "  $ " << n << " " << m << endl;
    for (int i = 0; m + i <= n; ++i) {
      ans[m + i] = m - i - 1;
      ans[m - i - 1] = m + i;
      mn = min(mn, m - i - 1);
    }
    n = mn - 1;
  }
  long long res = 0;
  n = n1;
  for (int i = 0; i <= n; ++i) {
    res += (i ^ ans[i]);
  }
  cout << res << endl;
  assert(res == 1ll * n * (n + 1));
  for (int i = 0; i <= n; ++i) {
    cout << ans[i] << " ";
  }
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
}
