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 <string.h>

char buff[200000+1];
int ans = 0;

int num(char c) { return c-'0'; }
int main() {
    scanf("%s", buff);
    //for (int i=4; i<200000; i++) buff[i]='0';
    //buff[200000]=0;
    int len = strlen(buff);

    for (int i=0; i+4<=len; i++) {
        int rest = num(buff[i])*100
            + num(buff[i+1])*10
            + num(buff[i+2]);
        printf("i=%d\n", i);
        for (int pos=i+3; pos<len; pos++) {
            //printf("rest=%d\n", rest);
            rest = rest*10+num(buff[pos]);
            rest %= 2019;
            if (rest==0) ans++;
        }
    }
    printf("%d\n", ans);

}