Why not find the mod directly with 6, as opposed to 2 and 3 ?

According to the constraints , the string length will not exceed 10e5. Here is the input for test case 5: 6966606919907090609037430091834960718075082030963209339303030899600689446690866466014339933090600886

wtf!!

include

include

char s[100005]; int arr[100005][3];

int fun(int x,int m){

if(x>=strlen(s))
return 0;
else if(arr[x][m]!=-1)
return arr[x][m];
else {
     if((s[x]-'0'+m)%3==0 && (s[x]-'0')%2==0){
    arr[x][m]=fun(x+1,(s[x]-'0'+m)%3)+1;
}
    else {
    arr[x][m]=fun(x+1,(s[x]-'0'+m)%3); }
    return arr[x][m];
}

}

int main(){ scanf(" %s",s); int j,i=0; long long ans=0; for(j=0;j<100005;j++){ arr[j][0]=arr[j][1]=arr[j][2]=-1;} while(s[i]!='\0'){ if((int)s[i]=='0'){

  ans=ans+1;
  }
else {
    ans=ans+fun(i,0);

    }

        i++;}
    printf("%lld",ans);

return 0;}
//can anyone please help out.
//my above code is giving time out for test cases above 12

https://www.hackerrank.com/contests/hourrank-18/challenges/super-six-substrings/submissions/code/1301177045

O(n) solution in python.. and I know nothing of dynamic programming

can anyone explain whats wrong with my code? public class superString {

public static void main(String[] args) {
    Scanner in = new Scanner(System.in);
    String s = in.next();
    int count = 0;
    for (int i = 0; i < s.length(); i++) {
        for (int j = 0; j <= s.length(); j++) {
            try {
                // System.out.println(s.substring(i,j));
                int k = Integer.valueOf(s.substring(i, j));
                if (k==0) {
                    count++;
                } else if ((s.substring(i, j).startsWith("0"))) {
                    count += 0;
                } else {

                    if (k % 6 == 0) {
                        count += 1;
                    }


                }
            }

            catch (Exception e) {

            }
        }

    }
    System.out.println(count);

}

}