/*
codr: timiter
task: mink
lang: C++
*/

#include <cstdio>

using namespace std;

int n;
int k[1024];

int main()
{
    int i, j, l;
    for (i = 0; i < 1020; i++)
    {
        k[i] = 0;
    }
    k[2] = 1;

    scanf("%d", &n);

    for (i = 3; i <= n; i++)
    {
        l = i;
        for (j = l / 2; j > 1 && l > 1; j--)
        {
            while (l % j == 0)
            {
                l /= j;
                k[i] += k[j];
            }
        }
        if (l > 1)
        {
            k[i] = i - 1;
        }
    }

    printf("%d\n", k[n]);

    return 0;
}
