/** n! means n * (n - 1) * ... * 3 * 2 * 1
For example, 10! = 10 * 9 * ... * 3 * 2 * 1 = 3628800,
and the sum of the digits in the number 10! is 3 + 6 + 2 + 8 + 8 + 0 + 0 = 27.
Find the sum of the digits in the number 100! */

#include "0.hpp"
#include "lib/lint.cpp"
#define _max 100

uint sum(string a)
{
  uint suma = 0;
  for(uint i = 0; i < a.size(); i++) suma += (a[i] - '0');
  return suma;
}

int main()
{
  lint a("1");
  lint b;
  for(int i = 1; i <= _max; i++) {
    b = itos(i);
    a = a * b;
  }
  
  cout << sum(a.getd()) << endl;
  
}