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 cast_spell(num):
  new_num = 0
  while num > 0:
    new_num += num % 10 # add the last digit
    num = num // 10     # get rid of the last digit
  return new_num

a = input()
i_count = 0
if len(a) > 1:
  # need to cast spell to string
  num = 0
  for ch in a:
    num += int(ch)   # add up the digit
  i_count += 1  # you just made one spell
  while num >= 10:
    num = cast_spell(num)
    i_count += 1    # every time you make the spell, increase the cout
print(i_count)
