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;
bool c[1000010];
int p[1000010], pw[21];
bool bit(int s, int i) { return 1 & (s >> i); }
int main() {
  int T, i, j, x, n, m, t;
  pw[0] = 1;
  for (i = 1; i <= 20; i++) pw[i] = pw[i - 1] << 1;
  while (~scanf("%d", &n)) {
    memset(c, true, sizeof(c));
    for (i = n; i >= 1; i--)
      if (c[i]) {
        int ans = 0, jj = 0;
        for (j = 20; j >= 0; j--)
          if (bit(i, j)) {
            jj = j;
            break;
          }
        for (j = jj; j >= 0; j--)
          if (bit(i, j) == 0) ans += pw[j];
        p[i] = ans;
        p[ans] = i;
        c[ans] = 0;
      }
    long long ans = 0;
    for (i = 0; i <= n; i++) ans += (p[i] ^ i);
    printf("%I64d\n", ans);
    for (i = 0; i <= n; i++) printf("%d ", p[i]);
    puts("");
  }
  return 0;
}
