CF1005D Polycarp and Div 3

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?

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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 $ .