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
import sys
import math
import copy
from collections import deque
from collections import *
from functools import lru_cache

input = lambda: sys.stdin.readline().rstrip("\r\n")


# sys.stdin = open('input.txt')
# sys.stdout = open('output.txt', 'w')

def ll(): return list(map(int, input().split()))


def lf(): return list(map(float, input().split()))


def ls(): return list(map(str, input().split()))


def mn(): return map(int, input().split())


def nt(): return int(input())


def ns(): return input()


def ip(n):
    if n % 2 == 0:
        return n == 2
    d = 3
    while d * d <= n and n % d != 0:
        d += 2
    return d * d > n


@lru_cache(None)
def daa(n):
    ct = 0
    for x in range(1, int(pow(n, 0.5)) + 1):
        if n % x == 0:
            ct += 1
            if x != n // x:
                ct += 1
    return ct
p=nt()

ct=0
while p>9:
    p=list(str(p))
    ans=0
    for x in p:
        ans+=int(x)
    p=ans
    ct+=1
print(ct)