Problem:
Score : 400 points

Problem Statement

Given is a string S consisting of digits from 1 through 9.

Find the number of pairs of integers (i,j) (1 ≤ i ≤ j ≤ |S|) that satisfy the following condition:

Condition: In base ten, the i-th through j-th characters of S form an integer that is a multiple of 2019.

Constraints

1 ≤ |S| ≤ 200000

S is a string consisting of digits from 1 through 9.

Input

Input is given from Standard Input in the following format:

S

Output

Print the number of pairs of integers (i,j) (1 ≤ i ≤ j ≤ |S|) that satisfy the condition.

Sample Input 1

1817181712114

Sample Output 1

3

Three pairs - (1,5), (5,9), and (9,13) - satisfy the condition.

Sample Input 2

14282668646

Sample Output 2

2

Sample Input 3

2119

Sample Output 3

0

No pairs satisfy the condition.

Solution in C (Clang 10.0.0) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<math.h>
int main(){
    int k=0,count=0;
    long long S;
    scanf("%lld",&S);
    while(pow(10,k)<=S){
        k++;
    }
    for(int i=1;i<k;i++){
        for(int j=i+3;j<=k;j++){
            long long s=S;
            long long sj=S/pow(10,j);
            s-=sj*pow(10,j);
            s/=pow(10,i-1);
            if(s%2019==0){
                count++;
            }
        }
    }
    printf("%d",count);
    return 0;
}