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 string alf = "abcdefghijklmnopqrstuvwxyz";
const int hashP = 239017;
const int N = 1e5 + 10;
const int MOD = 1e9 + 7;
const int MOD2 = 998244353;
const int INF = 1e9;
const long long INF2 = 1e18;
template <typename T>
bool umn(T &a, T b) {
  return (a > b ? (a = b, 1) : 0);
}
template <typename T>
bool umx(T &a, T b) {
  return (a < b ? (a = b, 1) : 0);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  string s = to_string(n);
  long long dl = s.size();
  long long ans = 0;
  for (long long mask = 1; mask <= (1 << dl); mask++) {
    long long ch = 0;
    vector<long long> a;
    for (int i = 0; i < dl; i++) {
      long long c = (mask >> i) & 1;
      a.push_back(c);
    }
    reverse(a.begin(), a.end());
    for (int i = 0; i < dl; i++) {
      ch *= 10;
      long long c = a[i];
      ch += c;
    }
    if (n >= ch) ans++;
  }
  cout << ans - 1;
  return 0;
}
