/*_______________ H E A D E R S ______________*/
#include <bits/stdc++.h>
#include <stdio.h>
#include <cmath>
/*_______________ M A C R O S ______________*/
#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(), v.rend()
#define pb push_back
#define ll long long int
#define yes cout << "YES\n"
#define no cout << "NO\n"
#define mx_ele(v) *max_element(v.begin(), v.end())
#define mnm_ele(v) *min_element(v.begin(), v.end())
#define endl "\n"
#define vint vector<int>
#define vll vector<long long int>
#define vstring vector<string>
#define pint vector<pair<int, int>>
#define pll vector<pair<ll, ll>>
#define fi first
#define se second
#define mp make_pair
#define Endl endl
#define mod abs
#define PIE 3.141592653589793238462643383279
using namespace std;
/* C O N V E R S I O N T O A N Y N U M B E R S Y S T E M ________
D E C I M A L T O A N Y R A D I X / B A S E ________*/
long long decimal_to_any(long long base, long long num)
{
if (num == 0)
{
return 0;
}
long long result = 0;
long long power = 1;
while (num > 0)
{
long long rem = num % base;
result += rem * power;
num /= base;
power *= 10;
}
return result;
}
/* -> HOW TO FIND THE NTH TERM OF SEQ <- ??
Solution:
-> formula : a1+(nth-1)*diff;
here,
>> a1 is the first element of the seq
>> nth is the term we want to find
>> diff is the difference of the sq
*/
ll nth_term(ll start_of_seq, ll diff, ll nth)
{
return start_of_seq + (nth - 1) * diff;
}
// G E T A L L T H E P E R M U T A T I O N O F A G I V E N S T R I N G
// RETURNS A VECTOR OF A STRING
vector<string> possible_permutation(string s)
{
sort(all(s));
vstring v;
do
{
v.pb(s);
} while (next_permutation(all(s)));
return v;
}
/* C O N V E R S I O N T O A N Y N U M B E R S Y S T E M ________
A N Y N U M B E R S Y S T E M T O D E C I M A L _______*/
ll any_to_decimal(ll base, string s)
{
reverse(all(s));
ll ans = 0;
for (int i = 0; i < s.size(); i++)
{
int digit = s[i] - '0';
ans += (digit * pow(base, i));
}
return ans;
}
// P O W E R O F T W O O R N O T ??
bool power_of_two(ll n)
{
if (n == 0)
return false;
return !(n & (n - 1));
}
// P R I M E O R N O T ??
bool isPrime(int n)
{
if (n < 2)
return 0;
for (int i = 2; i * i < n; i++)
if (n % i == 0)
return 0;
return 1;
}
long long numberOfDivisors(long long num)
{
long long total = 1;
for (int i = 2; (long long)i * i <= num; i++)
{
if (num % i == 0)
{
int e = 0;
do
{
e++;
num /= i;
} while (num % i == 0);
total *= e + 1;
}
}
if (num > 1)
{
total *= 2;
}
return total;
}
// D I G I T A L R O O T S !!
string digital_roots(string s)
{
while (s.size() > 1)
{
int ans = 0;
for (auto x : s)
{
ans += (x - '0');
}
s = to_string(ans);
}
return s;
}
// Prototype for cheems function
void cheems();
ll calc_trailing_zero(ll n)
{
ll ans = 0;
ll mul = 5;
while (n >= mul)
{
ans += (n / mul);
mul *= 5;
}
return ans;
}
// Bubble sort
/*
time complexity: O(N^2) [Quadratic]
space complexity: O(N) [Linear]
Total pass/steps: N-1
operation: (n*(n-1))/2;
the summation from 1-n is :n*(n+1)/2 and this is given by gauss's law.
*/
void bubble_sort(int v[], int n)
{
for (int i = 0; i < n - 1; i++)
{
cout << "\nSteps no " << i + 1 << ":" << endl;
for (int j = 0; j < n - i - 1; j++)
{
if (v[j] > v[j + 1])
{
swap(v[j], v[j + 1]);
}
for (int k = 0; k < n; k++)
{
cout << v[k] << ' ';
}
cout << endl;
}
}
}
// Binary search
/*
Pre_requisite: Sorting the array.
Time complexity : O(logN)
Space complexity: O(N)
*/
// M A I N F U N C T I O N !!
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
cout.tie(nullptr);
int _ = 1;
//cin >> _;
for (int i = 1; i <= _; i++)
{
cheems();
}
}
// V O I D C H E E M S !!
void cheems()
{
ll n;
cin >> n;
cout<<numberOfDivisors(n)<<endl;
}