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;
long long max(long long x, long long y) { return (x > y ? x : y); }
long long min(long long x, long long y) { return (x < y ? x : y); }
const long long maxn = 1000100, maxb = 23;
bool is[maxn];
long long ans[maxn];
long long HighBit(long long x) {
  x |= (x >> 1);
  x |= (x >> 2);
  x |= (x >> 4);
  x |= (x >> 8);
  x |= (x >> 16);
  x |= (x >> 32);
  return x - (x >> 1);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, r = 0;
  cin >> n;
  for (int i = n; i >= 0; i--) {
    if (is[i]) {
      continue;
    }
    long long tmp = HighBit(i) * 2 - (i + 1);
    tmp = (tmp < 0 ? 0 : tmp);
    is[i] = is[tmp] = 1;
    ans[tmp] = i;
    ans[i] = tmp;
  }
  for (int i = 0; i <= n; i++) {
    r += (ans[i] ^ i);
  }
  cout << r << endl;
  for (int i = 0; i < n + 1; i++) {
    cout << ans[i] << ' ';
  }
  return 0;
}
