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 mod = 1e9 + 7;
long long binPow(long long a, long long b) {
  long long x = 1, y = a;
  while (b) {
    if (b & 1) x = (x * y) % mod;
    y = (y * y) % mod;
    b >>= 1;
  }
  return x % mod;
}
long long inverserEuler(int n) { return binPow(n, mod - 2); }
long long C(int k, int n) {
  vector<long long> f(n + 1, 1);
  for (int i = 2; i <= n; i++) f[i] = (f[i - 1] * i) % mod;
  return (f[n] * ((inverserEuler(f[k]) * inverserEuler(f[n - k])) % mod) %
          mod) %
         mod;
}
long long x, y;
void extendEuclid(long long a, long long b) {
  if (b == 0) {
    x = 1;
    y = 0;
    return;
  }
  extendEuclid(b, a % b);
  long long x1 = y, y1 = x - (a / b) * y;
  x = x1;
  y = y1;
}
vector<long long> build(int n) {
  vector<long long> ans;
  int numOfDigit = log(n) / log(10) + 2, i, j;
  for (i = 0; i < numOfDigit + 1; ++i) {
    string num = string(i, '0') + string(numOfDigit - i, '1');
    do {
      ans.push_back(stoll(num));
    } while (next_permutation((num).begin(), (num).end()));
  }
  sort((ans).begin(), (ans).end());
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie();
  int n;
  cin >> n;
  vector<long long> allNum = build(n);
  int ans = 0;
  for (auto i : allNum) {
    if (i > 0 && i <= n)
      ans++;
    else if (i > n) {
      cout << ans;
      return 0;
    }
  }
  return 0;
}
