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 dp[12][12][5], len;
vector<long long int> vec;
void convert_digit(long long int n) {
  while (n > 0) {
    vec.push_back(n % 10);
    n = n / 10;
  }
}
long long int dynamic(long long int index, long long int value,
                      long long int tight) {
  if (index == -1) return 1;
  if (dp[index][value][tight] != -1) return dp[index][value][tight];
  long long int sum = 0;
  if (tight) {
    int mn = min(vec[index], (long long int)1);
    for (int i = 0; i <= mn; i++) {
      if (vec[index] == i) {
        sum = sum + dynamic(index - 1, i, 1);
      } else
        sum = sum + dynamic(index - 1, i, 0);
    }
  } else {
    for (int i = 0; i < 2; i++) {
      sum = sum + dynamic(index - 1, i, 0);
    }
  }
  return dp[index][value][tight] = sum;
}
int main() {
  long long int n;
  scanf("%lld", &(n));
  memset(dp, -1, sizeof(dp));
  convert_digit(n);
  long long int sum = 0;
  len = vec.size();
  for (int i = 0; i < 2; i++) {
    if (i == 1) {
      if (vec[len - 1] == 1)
        sum = sum + dynamic(len - 2, 1, 1);
      else
        sum = sum + dynamic(len - 2, 1, 0);
    } else
      sum = sum + dynamic(len - 2, 0, 0);
  }
  cout << sum - 1 << endl;
  return 0;
}
