D. Polycarp and Div 3

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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

Copy

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 6 forms 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 3333digits 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 numbers 00, 99, 201201 and 8181 are divisible by 33.

题意：任意分割这个数字字符串，使得各个部分被3整除的数量最大。

在计算数字字符串能否被3整除的方法是数位上数字之和是否被3整除。令分割起点为s，分割终点为e，当e-s+1>=3,那么必然这一段可以分割，比如像445....虽然4+4+5=13不被3整除，5也不被3整除，但是45可以啊，所以ans++;由于取余3的余数只有0,1,2，只要长度>=3时，必然可以分割了。

#include<bits/stdc++.h>
using namespace std;
int main()
{
    string a;
    int b[10];
    cin>>a;
    int n=a.size(),ans=0,cnt=0,sum=0;
    for(int i=0;i<n;i++)
    {
        b[cnt++]=a[i]-'0';
        sum+=b[cnt-1];
        if(sum%3==0||cnt>=3||b[cnt-1]%3==0)
        {
            ans++;
            cnt=0;
            sum=0;
        }
    }
    printf("%d\n",ans);
}