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 Rust (1.42.0) (judge verdict: Time Limit Exceeded)

use proconio::{input, marker::Chars};

fn main() {
    input! {
        s: Chars,
    }

    let len = s.len();

    if len < 4 {
        println!("0");
        return;
    }

    let mut count = 0;

    for i in 0..len-3 {
        // 先頭 3 つの数値の和を求める
        let mut sum = s[i..i+3].into_iter().map(|c| c.to_digit(10).unwrap()).sum::<u32>();
        let mut rest = s[i..i+3].into_iter().collect::<String>().parse::<u32>().unwrap();

        // j は i+3 から始める
        for j in i+3..len {
            let j_digit = s[j].to_digit(10).unwrap();
            sum = (sum + j_digit) % 3;
            rest = (rest * 10 + j_digit) % 673;

            if sum != 0 || rest != 0 {
                continue;
            }

            count += 1;
        }
    }

    println!("{}", count);
 }