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;
void boost() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
long long power(int x, int y) {
  long long ans = 1;
  while (y--) {
    ans = ans * x;
  }
  return ans;
}
int main() {
  boost();
  long long i, n, j;
  cin >> n;
  bitset<9> vs[512];
  for (i = 0; i < 9; i++) {
    long long ch = pow(2, i), co = 1;
    for (j = 0; j < 512; j++) {
      if (co <= ch) {
        vs[j][i] = 0;
        co++;
      } else {
        vs[j][i] = 1;
        co++;
        if (co > 2 * ch) co = 1;
      }
    }
  }
  vector<long long> v;
  for (i = 0; i < 512; i++) {
    long long ans = 0;
    for (j = 0; j < 9; j++) {
      if (!vs[i][j]) continue;
      ans = ans + vs[i][j] * power(10, j);
    }
    v.push_back(ans);
  }
  sort(v.begin(), v.end());
  v.push_back(1000000000);
  v.push_back(1010000000);
  auto it = upper_bound(v.begin(), v.end(), n);
  cout << (it - v.begin()) - 1;
}
