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
# # s = input()
# # ans = 0
# # q = s[::-1]
# # n = len(s) // 2

# # for i in range(n):
# #     if s[i] != q[i]:
# #         ans += 1

# # if ans == 1 or (ans == 0 and len(s) % 2 == 1):
# #     print("YES")
# # else:
# #     print("NO")


# a = int(input())
# l = list(map(int, input().split()))
# x = l.count(5)
# y = a - x

# if y == 0 :
#     print("-1")
# else:
#     x = (x // 9) * 9

#     if x == 0:
#         print("0")
#     else:
       
#         ans = '5' * x + '0' * y
#         print(ans)
def func(a):
    ans=0
    while a>0:
        r=a%10
        ans+=r
        #print(a)
        a=a//10
    return ans


a=int(input())
ans=0
while a>9:
    a=func(a)
    ans+=1
print(ans)
