G - Polycarp and Div 3

Polycarp likes numbers that are divisible by 3.

He has a huge number ss. 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 mm 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 33 that Polycarp can obtain?

Input

The first line of the input contains a positive integer ss. The number of digits of the number ss is between 1 and 2⋅105, 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.

Sample 1

Inputcopy	Outputcopy
3121	2

Sample 2

Inputcopy	Outputcopy
6	1

Sample 3

Inputcopy	Outputcopy
1000000000000000000000000000000000	33

Sample 4

Inputcopy	Outputcopy
201920181	4

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.

 

#include
#include
#include
using namespace std;
typedef long long ll;


int main()
{
	string a; cin >> a;
	int len = a.size();

	int sum = 0, num = 0, now = 0,ans = 0;
	for (int i = 0; i < len; i++)
	{
		now = a[i] - '0';
		sum += now;
		num++;
		//三种情况
		//第一种（num==3）：根据结论：一定能在一个长度大于或等于 33 的序列中划分出一个和能被 33 整除的子串。则将其划分成一组
		//第二种（now%3==3）：当前考虑的数是3的倍数，则将其划分成一组
		//第三种（sum%3==3）：当前和为3的倍数，则将其划分成一组
		if (num == 3|| now % 3 == 0 || sum % 3 == 0)
		{
			sum = 0;  //初始化
			num = 0;
			ans++; //总组数加一
		}
	}
	cout << ans << endl;
}