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 numbers 00, 99, 201201 and 8181 are divisible by 33.
题目大意：给你一串连续的数字让你找出能直接整除3的数，和相邻的可以凑成能被3整除的数，求他们的数量一共有多少。
看样列。
对于数字，他的每一位相加的和可以被三整除，那这个数就可以被三整除。
对于单个的可以被整除的，直接ans++就行了。
连续的两个数字，加起来能被三整除的话，也ans++;
连续的三个数字（前两个的和以及自身都不能被三整除的情况下），这三个数字中必然能凑出一个能被三整除的数字，或许是后两个，也或许是这三个一起。
why?
假如前两个凑不成被三整除的，那么他们对三取余的结果有两种，（1，1） （2 ，2）。为啥没有（1，2 )和 （2，1）？如果是这两种的话，那他俩加起来肯定能被三整除啊。所以只可能是（1，1）或者（2，2）这两种。
再来看第三个数字，假如他能被三整除，直接ans++，i++去看下一位（前两个死活也凑不成了，因为第三个自己就能被三整除）
假如他被三除余1，对于（1，1，1）和（2，2，1）这两种都行，ans++;
假如他被三除余2，对于（1，1，2）和（2，2，2）这两种也都行，ans++;
发现了什么？
在连续的三个数内，必然能找到至少一组正解。（前两个都不行的情况下，加上第三个一定能找到）
#include<iostream>
#include<string>
#include<cstring>
#include<cstdio>
using namespace std;
int main(){
string s;
while(cin>>s){
int n=s.length();
int ans=0;
int cnt=0;
int sum=0;
for(int i=0;i<n;i++){
sum+=s[i]-'0';
cnt++; //计数
if((s[i]-'0')%3==0||sum%3==0||cnt==3){
//本身能 或者 前2个能 或者 cnt==3
ans++;
sum=0;
cnt=0;
}
}
cout<<ans<<endl;
}
return 0;
}