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
import collections
import sys , threading
input = sys.stdin.readline


def main():

    n = input()
    l = len(n)
    memo = dict()
    def backtrack(index, prev, count):
        if index == l:
            return count
        new = prev + n[index]
        if int(new) % 3 == 0:
            ans = backtrack(index + 1, "", count + 1)
            memo[(index, prev)] = ans
            return ans
        elif int(n[index]) % 3 == 0:
            ans = backtrack(index + 1, "", count+1)
            memo[(index, prev)] = ans
            return ans
            
        else:
            ans1 = backtrack(index + 1, new, count)
            ans2 = backtrack(index + 1, n[index], count)
            memo[(index, prev)] = max(ans1, ans2)
            return max(ans1, ans2)
            
    backtrack(0,"", 0)
    print(memo[(0,"")])

sys.setrecursionlimit(1 << 30)
threading.stack_size(1 << 27)
main_thread = threading.Thread(target = main)
main_thread.start()
main_thread.join()
    