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?
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
3121
2
6
1
1000000000000000000000000000000000
33
201920181
4
4
0
10
1
11
0
12
1
13
1
31
1
14139582796
6
1670000
5
604500
5
40041
2
10000170
5
from collections import  defaultdict

bad=[]
for i in range(0,10**6,3):
    if len(str(i))==1:
        bad.append(str(i))
    else:
        s=str(i)
        fl=False
        for j in bad:
            if j in  s:
                fl=True
                break
        if not  fl:
            bad.append(s)

# bad=list(map(int,bad))

s=list(input())
res=""
ans=0
for id,i in enumerate(s):
    if i in bad:
        ans+=1
        continue
    if id+1<len(s):
        if i+s[id+1] in bad:
            ans+=1
            s[id+1]='.'
            continue
    if id+2<len(s):
        if i+s[id+1]+s[id+2] in bad:
            ans+=1
            s[id+1]="."
            s[id+1]="."
            continue
print(ans)



