#define _GLIBCXX_DEBUG
#include "bits/stdc++.h"
//#include <intrin.h>  //AtCoder (gcc) 上ではこれがあると動かない。__popcnt用のincludeファイル。
using namespace std;
using Graph = vector<vector<int>>;

typedef long long ll;
typedef long double ld;


#define int long long
#define rep(i, n) for(long long i = 0; i < (n); i++)
#define sqrt(d) pow((long double) (d), 0.50)

const ll INF = 2e9;
const ll large_P = 1e9 + 7;



//繰り返し2乗法
//N^aの、Mで割った余りを求める。
ll my_pow(ll N, ll a, ll M) {
	ll tempo;
	if (a == 0) {
		return 1;
	}
	else {
		if (a % 2 == 0) {
			tempo = my_pow(N, a / 2, M);
			return (tempo * tempo) % M;
		}
		else {
			tempo = my_pow(N, a - 1, M);
			return (tempo * N) % M;
		}
	}
}



//N_C_a を M で割った余り
ll my_combination(ll N, ll a, ll M) {
	ll answer = 1;

	rep(i, a) {
		answer *= (N - i);
		answer %= M;
	}

	rep(i, a) {
		answer *= my_pow(i + 1, M - 2, M);
		answer %= M;
	}

	return answer;
}



ll my_gcd(ll& a, ll& b) {
	if (b == 0) return a;
	ll temp = a % b;
	return my_gcd(b, temp);
}



// long long型 n を、base 進法での、i 桁目の数が入った vector にする。
void ll_to_vector(ll base, ll n, vector<ll>& v) {
	ll tempo = n;

	ll n_digit = floor(log10(n) / log10(base)) + 1;
	v.assign(n_digit, 0);

	//ll n_digit = v.size();

	for (ll i = 0; i < n_digit; i++) {
		v.at(i) = tempo / pow(base, n_digit - 1 - i);
		tempo -= v.at(i) * pow(base, n_digit - 1 - i);
	}
}



int char_to_int(char c) {
	switch (c) {
	case '0': return 0;
	case '1': return 1;
	case '2': return 2;
	case '3': return 3;
	case '4': return 4;
	case '5': return 5;
	case '6': return 6;
	case '7': return 7;
	case '8': return 8;
	case '9': return 9;
	default: return 0;
	}
}



signed main() {
	string S; cin >> S;
	int N = S.size();

	map<int, int> memo; memo[0]++;
	
	int amari = 0;
	int digit = 1;
	
	rep(i, N) {
		amari = (amari+ digit * char_to_int(S.at(N - 1 - i))) % 2019;
		digit = (digit * 10) % 2019;

		memo[amari]++;
		//cout << amari << endl;
	}

	int res = 0;
	for (auto&& iter = memo.begin(); iter != memo.end(); iter++) {
		res += iter->second * (iter->second - 1)/2;
		//cout << iter->first << " " << iter->second << endl;
	}

	cout << res << endl;
}