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 (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main (void) {
    char s[200001];
    scanf("%s", s);

    long long ans = 0;
    long long s_num;
    int s_len = strlen(s);

    for ( int i=0; i<s_len; i++ ) {
        s_num = strtol(&s[i], NULL, 10);
        while ( s_num > 0 && s_num >= 2019 ) {
            if ( s_num % 2019 == 0 )
                ans++;
            s_num /= 10;
        }
    }

    printf("%lld\n", ans);
    return EXIT_SUCCESS;
}