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?
The first line contains the only integer *n* (0 ≤ *n* ≤ 10<sup class="upper-index">100000</sup>). It is guaranteed that *n* doesn't contain any leading zeroes.
0
0
10
1
991
3
99
2
100
1
123456789
2
32
1
86
2
2
0
8
0
34
1
13
1
28
2
23
1
57
2
29
2
353869285
3
549548646
2
858893424
2
644818852
3
360322525
3
122937520
2
288403032
2
677257481
3
58059561
3
211288847
2
339900034079539584
2
784084029310269952
3
774730557264864000
3
478233807148352256
3
165646874056812544
3
477533739511673792
3
660119944377002240
3
448375218506078080
3
77213334785795616
3
165767221702271872
3
1
0
5
0
8
0
156161456146164104103460452045416165146141414651641564105461465156445621465146514562146145465165145145614561465145614561
3
9
0
def sum_of_digits(num):
    if num <= 9:
        return num, 0
    # else
    counter = 0 
    result = 0
    while num > 0:  # do not sum only 1 digit
        x = num % 10
        result += x
        num = num // 10
        counter += 1
    return result, counter

n = int(input())

# first case
x = sum_of_digits(n)
result = 0  # final output
# print(x)

summation = x[0]
count = x[1]
while count >= 1:  # more than one digit
    summation, count = sum_of_digits(summation)
    result += 1
    # print(count)

print(result)    