// yukicoder: No.278 連続する整数の和(2)
// 2019.4.14 bal4u

#include <stdio.h>
#include <math.h>

#define SIZE  50
long long factor[SIZE]; int power[SIZE];

int ptbl[] = {
		 3,   5,   7,  11,  13,  17,  19,  23,  29,
   31,  37,  41,  43,  47,  53,  59,  61,  67,  71,
   73,  79,  83,  89,  97, 101, 103, 107, 109, 113,
  127, 131, 137, 139, 149, 151, 157, 163, 167, 173,
  179, 181, 191, 193, 197, 199, 211, 223, 227, 229,
  233, 239, 241, 251, 257, 263, 269, 271, 277, 281,
  283, 293, 307, 311, 313, 317, 331, 337, 347, 349,
  353, 359, 367, 373, 379, 383, 389, 397, 401, 409,
  419, 421, 431, 433, 439, 443, 449, 457, 461, 463,
  467, 479, 487, 491, 499, 503, 509, 521, 523, 541,
  547, 557, 563, 569, 571, 577, 587, 593, 599, 601,
  607, 613, 617, 619, 631, 641, 643, 647, 653, 659,
  661, 673, 677, 683, 691, 701, 709, 719, 727, 733,
  739, 743, 751, 757, 761, 769, 773, 787, 797, 809,
  811, 821, 823, 827, 829, 839, 853, 857, 859, 863,
  877, 881, 883, 887, 907, 911, 919, 929, 937, 941,
  947, 953, 967, 971, 977, 983, 991, 997, 0 };

int prime_factor(long long n)
{
	int i, d, size;
	int *pp;

	size = 0;
	if ((n & 1) == 0) {
		factor[size] = 2;
		do n >>= 1, power[size]++;
		while ((n & 1) == 0);
		size++;
	}
	for (pp = ptbl; n > 1 && *pp > 0; pp++) {
		if (n % *pp) continue;
		d = *pp;
		factor[size] = d;
		do n /= d, power[size]++;
		while (n % d == 0);
		size++;
	}
	if (n > 1) {
		int b = (int)sqrt((double)n);
		for (i = 1009; n > 1; i += 2) {
			if (i > b) {
				factor[size] = n, power[size++] = 1;
				break;
			}
			if (n % i == 0) {
				factor[size] = i;
				do n /= i, power[size]++;
				while (n % i == 0);
				size++;
			}
		}
	}
	return size;
}

long long mypow(long long x, int p)
{
	long long r = 1;
	while (p) {
		if (p & 1) r = r * x;
		x = x * x;
		p >>= 1;
	}
	return r;
}

int main()
{
	int i, sz;
	long long N, ans;

	scanf("%lld", &N);
	if ((N & 1) == 0) N >>= 1;
	sz = prime_factor(N);

	ans = 1;
	for (i = 0; i < sz; i++) {
		if (power[i] == 1) ans *= factor[i] + 1;
		else ans *= (mypow(factor[i], power[i] + 1) - 1) / (factor[i] - 1);
	}
	printf("%lld\n", ans);
	return 0;
}