// i try to solve this problem ^^ if i wrong, dont laugh me :)
// We easy see that if n = 1 we dont need to cut and ans is 0
// if n = 2 we need 1 cut
// i call a[i] is the number of cut to make cake to i people.
// now if n is a odd number we need 1 cut to make the cake in 2 side (n-1)/n and 1/n
// now we need to share (n-1)/n cake to n-1 people (n-1) now is even. we habe a[n] = a[n-1] + 1;
// if n is a even number, we need 1 cut to make the cake in 2 same side so a[n] = a[n/2] + 1;

// It's code

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <vector>
#include <cstring>
#include <cmath>

using namespace std;

int n, m, i, j, a[100009];

int main() {
    cin >> n;                       // cin n people
    a[1] = 0;                       // if n = 1 we need 0 cut
    a[2] = 1;                       // if n = 2 we need 1 cut
    for(i = 3; i <= n; i++){        // make a sequence a[i] is the answer of i people
        if(i % 2 == 1){             // if people is odd
            a[i] = a[i - 1] + 1;
        }
        else{                       // if people is even
            a[i] = a[i / 2] + 1;
        }
    }
    cout << a[n] << "\n";
}