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;
const int N = 4e3 + 5;
const int MAX = 1e9 + 7;
long long n;
void boost() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
int main() {
  boost();
  cin >> n;
  long long k = 0ll;
  for (long long x = 0ll; x < 2ll; x++) {
    for (long long a = 0ll; a < 2ll; a++) {
      for (long long b = 0ll; b < 2ll; b++) {
        for (long long c = 0ll; c < 2ll; c++) {
          for (long long d = 0ll; d < 2ll; d++) {
            for (long long e = 0ll; e < 2ll; e++) {
              for (long long f = 0ll; f < 2ll; f++) {
                for (long long g = 0ll; g < 2ll; g++) {
                  for (long long h = 0ll; h < 2ll; h++) {
                    for (long long i = 0ll; i < 2ll; i++) {
                      if (a + 10ll * b + c * 100ll + d * 1000ll + e * 10000ll +
                                  f * 100000ll + g * 1000000ll +
                                  h * 10000000ll + i * 100000000ll +
                                  x * 1000000000ll <=
                              n &&
                          a + 10ll * b + c * 100ll + d * 1000ll + e * 10000ll +
                                  f * 100000ll + g * 1000000ll +
                                  h * 10000000ll + i * 100000000ll +
                                  x * 1000000000ll >=
                              1ll) {
                        k++;
                      }
                    }
                  }
                }
              }
            }
          }
        }
      }
    }
  }
  cout << k;
  return 0;
}
