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 33. 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+1m+1 parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by 33.

For example, if the original number is s=3121s=3121, then Polycarp can cut it into three parts with two cuts: 3|1|213|1|21. As a result, he will get two numbers that are divisible by 33.

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 11 and 2⋅1052⋅105, inclusive. The first (leftmost) digit is not equal to 0.

Output

Print the maximum number of numbers divisible by 33 that Polycarp can get by making vertical cuts in the given number ss.

Examples

Input

3121

Output

2

Input

6

Output

1

Input

1000000000000000000000000000000000

Output

33

Input

201920181

Output

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 6forms one number that is divisible by 33.

In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and 3333 digits 0. Each of the 3333 digits 0 forms a number that is divisible by 33.

In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers00, 99, 201201 and 8181 are divisible by 33.

这套题可能和我五行不合，比较克我，这道题补题的时候也过了，题意是你看这个字符串当中有多少个3的倍数，然后你看代码，我把解释都写进去了。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
char s[200100];
int main()
{
    while(~scanf("%s",s))
        {
            int l=strlen(s);//定义字符串的长度 
            int num=0,sum=0;
            int a,b=0;
            for(int i=0; i<l; i++)
                {
                    a=s[i]-'0';
                    if(a%3==0)//有一个3就++； 
                        num++;
                }
            for(int i=0; i<l; i++)
                {
                    a=s[i]-'0';
                    if(a%3!=0)
                        {
                            b=b*10+a;//因为你第一遍扫了一遍，第二次你就累加求和，看看他能不能加成3的倍数 
                            sum++;
                            if(sum==3||b%3==0)//这个sum==3是因为当有三个都不是3的倍数相加的话，那么他一定是3的倍数。 
                            {
                                num++;
                                b=0;//统计完了就进行清0，小心出问题。 
                                sum=0;
                            }
                        }
                    else
                        {
                            b=0;
                            sum=0;
                        }
                }
            printf("%d\n",num);//输出统计的个数 
        }
    return 0;
}