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;
struct trie {
  trie *next[2];
  int how;
  trie() {
    how = 0;
    for (int i = 0; i < 2; i++) next[i] = NULL;
  }
  ~trie() {
    for (int i = 0; i < 2; i++)
      if (next[i] != NULL) delete next[i];
  }
};
void insert(trie *root, long long num) {
  trie *curr = root;
  for (int i = 30; i >= 0; i--) {
    if (!curr->next[bool(num & 1LL << i)])
      curr->next[bool(num & 1LL << i)] = new trie;
    curr = curr->next[bool(num & 1LL << i)];
    curr->how++;
  }
}
void del(trie *root, long long num) {
  trie *curr = root;
  for (int i = 30; i >= 0; i--) {
    trie *temp = curr;
    curr = curr->next[bool(num & 1LL << i)];
    curr->how--;
    if (!curr->how) temp->next[bool(num & 1LL << i)] = NULL;
  }
}
long long match(trie *root, long long num) {
  trie *curr = root;
  long long res = 0;
  for (int i = 30; i >= 0; i--) {
    if (curr->next[bool(num & 1LL << i)]) {
      curr = curr->next[bool(num & 1LL << i)];
      if (bool(num & 1LL << i)) res |= 1LL << i;
    } else if (curr->next[!bool(num & 1LL << i)]) {
      curr = curr->next[!bool(num & 1LL << i)];
      if (!bool(num & 1LL << i)) res |= 1LL << i;
    }
  }
  return res;
}
trie root;
int ans[2000000];
int main() {
  ios_base::sync_with_stdio(0);
  int n;
  cin >> n;
  for (int i = 0; i <= n; i++) {
    insert(&root, i);
  }
  for (int i = n; i >= 0; i--) {
    ans[i] = match(&root, i ^ ((1LL << 30) - 1));
    del(&root, ans[i]);
  }
  long long res = 0;
  for (int i = 0; i <= n; i++) res += ans[i] ^ i;
  cout << res << endl;
  for (__typeof(n + 1) i = 0; i < (n + 1); i++) {
    if (i) cout << " ";
    cout << ans[i];
  }
  cout << endl;
  return 0;
}
