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.

Examples

Input

1817181712114


Output

3


Input

14282668646


Output

2


Input

2119


Output

0
import java.util.Scanner;


public class Main {

    public static String Y = "Yes";
    public static String N = "No";
    public static String I = "IMPOSSIBLE";
    public static String B = "Bad";
    public static String G = "Good";
    public static long MOD = (long) (Math.pow(10, 9) + 7);
    public static Scanner sc = new Scanner(System.in);

    public static void main(String[] args) {

        String s = n();
        int n = s.length();
        int[] mod = new int[2019];
        mod[0]++;

        int keta = 1;
        int top = 0;
        int now = 0;

        for (int i = n - 1; i >= 0; i--) {
            top = s.charAt(i) - '0';

            now = top * keta + now;
            now %= 2019;
            mod[now]++;

            keta *= 10;
            keta %= 2019;
        }

        long ans = 0l;
        for (int count : mod) {
            if (count > 1) {
                count--;
                ans += count * (count + 1) / 2l;
            }
        }
        out(ans);

    }

    public static int count(String search, String text) {

        int nsearch = search.length();
        int ntext = text.length();
        int count = 0;

        for (int i = 0; i <= ntext - nsearch + 1; i++) {
            for (int j = i + nsearch - 1; j < ntext; j++) {
                if (search.equals(text.substring(i, j + 1))) {
                    count++;
                }
            }
        }

        return count;
    }

    /*
     * 以下メソッド集
     */
    static int ni() {

        return sc.nextInt();
    }

    static long nl() {

        return sc.nextLong();
    }

    static String n() {

        return sc.next();
    }

    static char[] nc() {

        return sc.next().toCharArray();
    }

    static int kaijo(int n) {

        if (n == 0 || n == 1) {
            return 1;
        } else {
            return n * kaijo(n - 1);
        }
    }

    static int fib(int n) {

        return (n == 1 || n == 0) ? n : fib(n - 2) + fib(n - 1);
    }

    static long lcm(long m, long n) {

        return m * n / gcd(m, n);
    }

    static int lcm(int m, int n) {

        return m * n / gcd(m, n);
    }

    static long gcd(long a, long b) {

        return (b == 0) ? a : gcd(b, a % b);
    }

    static int gcd(int a, int b) {

        return (b == 0) ? a : gcd(b, a % b);
    }

    static void out(String info) {

        System.out.println(info);
    }

    static void out(int info) {

        System.out.println(info);
    }

    static void out(double info) {

        System.out.println(info);
    }

    static void out(long info) {

        System.out.println(info);
    }

    static void out(char info) {

        System.out.println(info);
    }

    static void outn(String info) {

        System.out.print(info);
    }

    static void outn(int info) {

        System.out.print(info);
    }

    static void outn(double info) {

        System.out.print(info);
    }

    static void outn(long info) {

        System.out.print(info);
    }

    static void outn(char info) {

        System.out.print(info);
    }

    static int max(int a, int b) {

        return Math.max(a, b);
    }

    static long max(long a, long b) {

        return Math.max(a, b);
    }

    static int min(int a, int b) {

        return Math.min(a, b);
    }

    static long min(long a, long b) {

        return Math.min(a, b);
    }
}