take n, multiply the digits until only 1 digit remains and write the number of iterations

#include <iostream>
using namespace std;

int main()
{
  int a,b,c,d;
  int sum = 1;
  int sum1 = 1;
  int sum2 = 1;
  int sum3 = 1;
  int br = 0;
  int number;
  cin >> number;
  if(number >= 0 && number <= 9)
  {
    cout << br <<  endl;
    return 0;
  }
  while(number > 0)
  {
        a = number % 10;
        sum = sum * a;
        number = number / 10;
    }
  br = br + 1;
  if(sum > 9)
  {
  while(sum > 0)
  {
   b = sum % 10;
   sum1 = sum1 * b;
   sum = sum / 10;
    }
    }else{
    cout << br << endl;
    return 0;
    }
  br = br + 1;
  if(sum1 > 9)
  {
    while(sum1 > 0)
  {
      d = sum1 % 10;
      sum2 = sum2 * d;
      sum1 = sum1 / 10;
    }
  }else{
    cout << br << endl;
    return 0;
  }
   br = br + 1;
  if(sum2 > 9)
  {
    while(sum2 > 0)
      {
        c = sum2 % 10;
        sum3 = sum3 * c;
        sum2 = sum2 / 10;
      }
  }else{
    cout << br << endl;
    return 0;
  }
  br = br + 1;
  cout << br << endl;
}
/*
39 --> 3 (because 3*9 = 27, 2*7 = 14, 1*4 = 4 and 4 has only one digit)
999 --> 4 (because 9*9*9 = 729, 7*2*9 = 126, 1*2*6 = 12, and finally 1*2 = 2)
4 --> 0 (because 4 is already a one-digit number)
*/