Problem: Polycarp likes numbers that are divisible by 3.

He has a huge number s. Polycarp wants to cut from it the maximum number of numbers that are divisible by 3. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after m such cuts, there will be m+1 parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by 3.

For example, if the original number is s=3121, then Polycarp can cut it into three parts with two cuts: 3|1|21. As a result, he will get two numbers that are divisible by 3.

Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.

What is the maximum number of numbers divisible by 3 that Polycarp can obtain?

Input

The first line of the input contains a positive integer s. The number of digits of the number s is between 1 and 2⋅10^5, inclusive. The first (leftmost) digit is not equal to 0.

Output

Print the maximum number of numbers divisible by 3 that Polycarp can get by making vertical cuts in the given number s.

Examples

Input

3121


Output

2


Input

6


Output

1


Input

1000000000000000000000000000000000


Output

33


Input

201920181


Output

4

Note

In the first example, an example set of optimal cuts on the number is 3|1|21.

In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by 3.

In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and 33 digits 0. Each of the 33 digits 0 forms a number that is divisible by 3.

In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers 0, 9, 201 and 81 are divisible by 3.

Solution (in PYTHON3):

Solution:
# Hey, there Stalker!!!
# This Code was written by:
# ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
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# ▒▒▒▒▒▒▒▒▒▒▒╔╝║▒▒▒▒▒▒▒
# ▒▒▒▒▒▒▒▒▒▒▒╚═╝▒▒▒▒▒▒▒
# ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
#from functools import reduce
from __future__ import division, print_function
#mod=int(1e9+7)
#import resource
#resource.setrlimit(resource.RLIMIT_STACK, [0x100000000, resource.RLIM_INFINITY])
#import threading
#threading.stack_size(2**26)
"""fact=[1]
#for i in range(1,100001):
#    fact.append((fact[-1]*i)%mod)
#ifact=[0]*100001
#ifact[100000]=pow(fact[100000],mod-2,mod)
#for i in range(100000,0,-1):
#    ifact[i-1]=(i*ifact[i])%mod"""
#from collections import deque, defaultdict, Counter, OrderedDict
#from math import ceil, sqrt, hypot, factorial, pi, sin, cos, radians, gcd
#from heapq import heappush, heappop, heapify, nlargest, nsmallest


# sys.setrecursionlimit(10**6) 
from sys import stdin, stdout
import bisect            #c++ upperbound
from bisect import bisect_left as bl              #c++ lowerbound bl(array,element)
from bisect import bisect_right as br             #c++ upperbound
import itertools
from collections import Counter
import collections
import math
import heapq
import re
def modinv(n,p):
    return pow(n,p-2,p)
def cin():
    return map(int,sin().split())
def ain():                           #takes array as input
    return list(map(int,sin().split()))
def sin():
    return input()
def inin():
    return int(input())
  
def Divisors(n) : 
    l = []  
    for i in range(1, int(math.sqrt(n) + 1)) :
        if (n % i == 0) : 
            if (n // i == i) : 
                l.append(i) 
            else : 
                l.append(i)
                l.append(n//i)
    return l

def most_frequent(list):
    return max(set(list), key = list.count) 
def GCD(x,y):
    while(y):
        x, y = y, x % y
    return x

def ncr(n,r,p):                                              #To use this, Uncomment 19-25 
    t=((fact[n])*((ifact[r]*ifact[n-r])%p))%p
    return t

def Convert(string): 
    li = list(string.split("")) 
    return li 

def SieveOfEratosthenes(n): 
    global prime
    prime = [True for i in range(n+1)] 
    p = 2
    while (p * p <= n): 
        if (prime[p] == True): 
            for i in range(p * p, n+1, p): 
                prime[i] = False
        p += 1
    f=[]
    for p in range(2, n): 
        if prime[p]: 
            f.append(p)
    return f
prime=[]
q=[]
def dfs(n,d,v,c):
    global q
    v[n]=1
    x=d[n]
    q.append(n)
    j=c
    for i in x:
        if i not in v:
            f=dfs(i,d,v,c+1)
            j=max(j,f)
            # print(f)
    return j

#Implement heapq
#grades = [110, 25, 38, 49, 20, 95, 33, 87, 80, 90] 
#print(heapq.nlargest(3, grades)) #top 3 largest
#print(heapq.nsmallest(4, grades))
#Always make a variable of predefined function for  ex- fn=len
#n,k=map(int,input().split())
"""*******************************************************"""
def main():
    a=list(map(int,str(input())))
    count=0
    s=0
    ans=0
    temp=[]
    for i in range(len(a)):
        if a[i]%3==0:
            ans+=1
            s=0
            count=0
        else:
            if a[i]%3==1:
                s+=1
            else:
                s+=2
            count+=1
            if s%3==0 or count==3:
                count=0
                s=0
                ans+=1

    print(ans)
























######## Python 2 and 3 footer by Pajenegod and c1729
py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip
    range = xrange
 
import os, sys
from io import IOBase, BytesIO
 
BUFSIZE = 8192
class FastIO(BytesIO):
    newlines = 0
 
    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write if self.writable else None
 
    def _fill(self):
        s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
        self.seek((self.tell(), self.seek(0,2), super(FastIO, self).write(s))[0])
        return s
 
    def read(self):
        while self._fill(): pass
        return super(FastIO,self).read()
 
    def readline(self):
        while self.newlines == 0:
            s = self._fill(); self.newlines = s.count(b"\n") + (not s)
        self.newlines -= 1
        return super(FastIO, self).readline()
 
    def flush(self):
        if self.writable:
            os.write(self._fd, self.getvalue())
            self.truncate(0), self.seek(0)
 
class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        if py2:
            self.write = self.buffer.write
            self.read = self.buffer.read
            self.readline = self.buffer.readline
        else:
            self.write = lambda s:self.buffer.write(s.encode('ascii'))
            self.read = lambda:self.buffer.read().decode('ascii')
            self.readline = lambda:self.buffer.readline().decode('ascii')
 
 
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')
if __name__== "__main__":
    main()
#threading.Thread(target=main).start()