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;
long long int M = 1e9 + 7;
long long int inf = 9 * 1e18;
long long binpow(long long val, long long deg) {
  if (deg < 0) return 0;
  if (!deg) return 1 % M;
  if (deg & 1) return binpow(val, deg - 1) * val % M;
  long long res = binpow(val, deg >> 1);
  return (res * res) % M;
}
long long modinv(long long n) { return binpow(n, M - 2); }
bool sortbysec(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.second > b.second);
}
long long countDigit(long long n) {
  long long count = 0;
  while (n != 0) {
    n = n / 10;
    ++count;
  }
  return count;
}
void solve() {
  long long n;
  cin >> n;
  long long m = 111111111;
  long long ans = 0;
  while (m > 0) {
    while (n / m == 0) {
      m /= 10;
    }
    long long c = countDigit(m);
    ans += pow(2, c) - 1;
    long long m2 = pow(10, c);
    if (n / m2 == 0) {
      cout << ans << "\n";
      return;
    } else if (n % m2 == 0) {
      ans++;
      cout << ans << "\n";
      return;
    } else {
      ans++;
      n %= m2;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
