/*
TASK: sum
LANG: C++
*/

#include <stdio.h>
#include <cmath>

using namespace std;

long n;
int br = 0;

int main () {
    scanf ("%ld", &n);
    double k = (double) sqrt(n);
    long l = n / 2;
    if (n == 3) {
        printf("1\n");
        return 0;
    }
    for (int i = 2; i <= l; i ++) {
        if (double(n) / i < i / 2.) break;
        if (i % 2) {
            if (n % i == 0) {
                br ++;
            }
        } else {
            int m = i, evenbr = 1;
            while (m % 2 == 0) {
                m /= 2;
                evenbr <<= 1;
            }
            evenbr >>= 1;
            if ((double(n) / i) - n / i == double(evenbr) / i) {
                br ++;
            }
        }
    }
    printf ("%d\n", br);
    return 0;
}
