One beautiful July morning a terrible thing happened in Mainframe: a mean virus Megabyte somehow got access to the memory of his not less mean sister Hexadecimal. He loaded there a huge amount of n different natural numbers from 1 to n to obtain total control over her energy.

But his plan failed. The reason for this was very simple: Hexadecimal didn't perceive any information, apart from numbers written in binary format. This means that if a number in a decimal representation contained characters apart from 0 and 1, it was not stored in the memory. Now Megabyte wants to know, how many numbers were loaded successfully.

Input

Input data contains the only number n (1 ≤ n ≤ 109).

Output

Output the only number — answer to the problem.

Examples

Input

10


Output

2

Note

For n = 10 the answer includes numbers 1 and 10.
#include <bits/stdc++.h>
using namespace std;
string n;
map<pair<string, int>, int> memo;
bool valid(string ss) {
  if (ss.size() < n.size()) return true;
  if (ss.size() == n.size() && ss <= n) return true;
  return false;
}
int bktk(string s, int rem) {
  if (s.size() >= n.size()) return 0;
  if (memo[make_pair(s, rem)] != 0) return memo[make_pair(s, rem)];
  int c1 = 0, c2 = 0;
  if (valid(s + '1')) c1 = bktk(s + '1', rem - 1) + 1;
  if (valid(s + '0')) c2 = bktk(s + '0', rem - 1) + 1;
  return memo[make_pair(s, rem)] = c1 + c2;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n;
  cout << bktk("1", n.size()) + 1;
}
