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 n, i;
bool powersOf2[1500001];
int luat[1500001];
long long sum = 0;
vector<int> sol;
int main() {
  cin >> n;
  for (i = 1; i <= n + n; i *= 2) {
    powersOf2[i] = true;
  }
  for (i = 0; i <= n; i++) {
    luat[i] = -1;
  }
  for (i = n; i >= 0; i--) {
    if (luat[i] > -1) {
      sum += (long long)(luat[i] ^ i);
      sol.push_back(luat[i]);
      continue;
    }
    int mask = 1;
    int cmask = ~i;
    int lastcmask = -1;
    int contor = 0;
    while (!(((cmask & mask) >= 0) && ((cmask & mask) <= i)) && contor < 32) {
      mask <<= 1;
      mask |= 1;
      contor++;
    }
    while (((cmask & mask) >= 0) && ((cmask & mask) <= i) && contor < 32) {
      if (luat[(cmask & mask)] == -1) lastcmask = cmask & mask;
      mask <<= 1;
      mask |= 1;
      contor++;
    }
    if (lastcmask != -1) {
      luat[lastcmask] = i;
      sum += (long long)(lastcmask ^ i);
      sol.push_back(lastcmask);
    } else {
      sum += (long long)i;
      sol.push_back(i);
    }
  }
  cout << sum << "\n";
  for (i = sol.size() - 1; i >= 0; i--) {
    cout << sol[i] << " ";
  }
}
