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 fac(long long n) {
  long long m = 1;
  for (long long i = 1; i <= n; i++) {
    m = m * i;
  }
  return m;
}
long long gcd(long long x, long long y) {
  if (x > y) swap(x, y);
  long long md;
  while (1) {
    md = y % x;
    if (md == 0) break;
    y = x;
    x = md;
  }
  return x;
}
long long nCr(long long n, long long r) {
  long long rs = (fac(n)) / (fac(r) * fac(n - r));
  return rs;
}
long long nPr(long long n, long long r) {
  long long rs = fac(n) / fac(n - r);
  return rs;
}
bool flag, fl;
long long i, j;
long long pw(long long n, long long p) {
  long long mn = 1;
  if (p == 0) return 1;
  for (long long z = 0; z < p; z++) {
    mn *= n;
  }
  return mn;
}
int main() {
  long long n;
  scanf("%lld", &n);
  long long num = 0;
  vector<int> bn;
  long long rsl;
  for (i = 1; i <= 1024; i++) {
    bn.clear();
    j = i;
    num = 0;
    while (j > 0) {
      bn.push_back(j % 2);
      j = j / 2;
    }
    long long ln = bn.size();
    for (j = 0; j < ln; j++) {
      num += (bn[j] * pw(10, j));
    }
    if (num <= n) {
      rsl = i;
    } else
      break;
  }
  printf("%lld\n", rsl);
  return 0;
}
