Your task is to calculate the number of trailing zeroes in the factorial **n!**

for example, **20! = 2432902008176640000**

# Input

The only input line has an integer **n**.

# Output

Print the number of trailing zeros in **n**!

# Constraints

**1 <= n <= 10^9**

# Example

# Input:

```
20
```

Output:

```
4
```

To solve this problem, we just need to count the number of **5's** and number of **2s** in the pr**i**me factorization of the number **n.**

We can see that the ending digit of the given number is **0** which is obtained only by multiplying the given number **5** with **2.**

Hence, for the number of factors given number **n**, we can add them up and that will be our answer

Let's code the solution up.

```
#include<bits/stdc++.h>
using namespace std;
int getFactors(int n){
int cnt = 0;
while(n % 5 == 0)
{
n /= 5;
cnt++;
}
return cnt;
}
int primeFactrization(int n)
{
int ret = 0;
for(int i = 5; i <= n; i += 5)
{
ret += getFactors(i);
}
return ret;
}
int main(){
int n;
cin >> n;
cout << primeFactorization(n);
}
```

That's all we have to do in this problem, I shall see you soon in the next blog.