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, m, ans[1000005];
bool mark[1000005];
bool has(int a, int b) { return (a & (1 << b)); }
bool submask(int a, int b) {
  for (int i = 0; i < 20; i++) {
    if (has(b, i) && !has(a, i)) return false;
  }
  return true;
}
int main() {
  while (scanf("%d", &n) != EOF) {
    for (int i = 0; i <= n; i++) ans[i] = i;
    int mask = 1;
    while (mask <= n) mask <<= 1;
    mask -= 1;
    memset(mark, false, sizeof(mark));
    long long sum = 0;
    for (int i = n; i >= 0; i--) {
      if (mark[i]) continue;
      while ((mask ^ i) > n || mark[mask ^ i]) mask >>= 1;
      ans[i] = mask ^ i;
      ans[mask ^ i] = i;
      sum += mask * 2;
      mark[i] = true, mark[mask ^ i] = true;
      ;
    }
    printf("%I64d\n", sum);
    for (int i = 0; i <= n; i++) {
      if (i) printf(" ");
      printf("%d", ans[i]);
    }
    printf("\n");
  }
  return 0;
}
