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?
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
3121
2
6
1
1000000000000000000000000000000000
33
201920181
4
4
0
10
1
11
0
12
1
13
1
31
1
14139582796
6
1670000
5
604500
5
40041
2
10000170
5
#include <bits/stdc++.h>
#define int long long
#define all(v) v.begin(),v.end()
#define MOD (int)(1e9+7)

using namespace std;

void solve() {
    string str;
    int res = 0, j = 0, sum = 0;
    cin >> str;
    for(int i = 0; i < str.length(); i++) {
        int temp = str[i] - 48;
        temp %= 3;
        sum += temp;
        j++;
        if(sum % 3 == 0 || temp == 0 || j == 3) {
            res++;
            sum = 0;
            j = 0;
        }
    }
    cout << res;
}

signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);cout.tie(nullptr);
    solve(); 
}








