## Algorithm

B. Sum of Digits
time limit per test
2 seconds
memory limit per test
265 megabytes
input
standard input
output
standard output

Having watched the last Harry Potter film, little Gerald also decided to practice magic. He found in his father's magical book a spell that turns any number in the sum of its digits. At the moment Gerald learned that, he came across a number n. How many times can Gerald put a spell on it until the number becomes one-digit?

Input

The first line contains the only integer n (0 ≤ n ≤ 10100000). It is guaranteed that n doesn't contain any leading zeroes.

Output

Print the number of times a number can be replaced by the sum of its digits until it only contains one digit.

Examples
input
Copy
`0`
output
Copy
`0`
input
Copy
`10`
output
Copy
`1`
input
Copy
`991`
output
Copy
`3`
Note

In the first sample the number already is one-digit — Herald can't cast a spell.

The second test contains number 10. After one casting of a spell it becomes 1, and here the process is completed. Thus, Gerald can only cast the spell once.

The third test contains number 991. As one casts a spell the following transformations take place: 991 → 19 → 10 → 1. After three transformations the number becomes one-digit.

## Code Examples

### #1 Code Example with C++ Programming

```Code - C++ Programming```

``````#include <cstdio>
#include <iostream>

long sumOfDigits(long x){
long output(0);
while(x > 0){output += x%10; x /= 10;}
return output;
}

int main(){

std::string input("");getline(std::cin, input);
if(input.length() == 1){puts("0");}
else{
long current(0), times(1);
for(long k = 0; k < input.length(); k++){current += input[k] - '0';}
while(current > 9){current = sumOfDigits(current); ++times;}
printf("%ld\n", times);
}
return 0;
}``````
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