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 functools import lru_cache
import sys

input = sys.stdin.readline

num = input().strip()

n = len(num)

# @lru_cache(None)
# def dp(i, sum_under_mod):
#     if i >= n:
#         return 0
    
#     sum_under_mod = ( sum_under_mod + int(num[i]) ) % 3
#     return max(
#         dp(i + 1,  sum_under_mod),
#         dp(i + 1, 0) + (sum_under_mod % 3 == 0)
#     )

# print(dp(0, 0))

dp = [[0 for _ in range(3)] for _ in range(n + 1)]

for i in range(n - 1, -1, -1):
    for j in range(2, -1, -1):
        hold = (j + int(num[i])) % 3
        dp[i][j] = max(dp[i + 1][hold], dp[i + 1][0] + (hold % 3 == 0))

print(dp[0][0])
# print(dp)

# 3 1 2 1
# 6