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

2105

, 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

Copy

Output

Copy

Input

Copy

Output

Copy

Input

Copy

Output

Copy

Input

Copy

Output

Copy

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

其实并没有dp直接贪心即可

我们考虑分类讨论即可 我在比赛中被hack了 因为我没有判断最后一个是什么

讨论情况1、遇到0一定切刀

2、如果前后两个非0且不一样一定切刀

3、 如果前后两个相同则一定需要切刀 因为一定会存在满足条件的情况

注意最后的边界特判

具体讨论细节看代码即可


 

分类: 贪心

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