// Original C++ solution (slower):
/*

    Author : Nishant Gupta 2.0

*/

#include<bits/stdc++.h>



#define LL long long int

#define getcx getchar_unlocked

#define X first

#define Y second

#define PB push_back

#define MP make_pair

#define MAX 200005

#define MOD 1000000007

#define INF 0x3f3f3f3f

#define INFL (LL(1e18))

#define rep(i, a, n) for(i=a;i<n;i++)

#define rev(i, a, n) for(i=a;i>=n;i--)

#define MSV(A,a) memset(A, a, sizeof(A))

#define rep_itr(itr, c) for(itr = (c).begin(); itr != (c).end(); itr++)



using namespace std;



inline void inp( int &n ) { 

	n=0;    int ch=getcx();int sign=1;   

	while( ch < '0' || ch > '9' ){if(ch=='-')sign=-1; ch=getcx();} 

	while(  ch >= '0' && ch <= '9' )    

	n = (n<<3)+(n<<1) + ch-'0', ch=getcx();

	n=n*sign;

}



inline void inp1( LL &n ) { 

	n=0;    LL ch=getcx();LL sign=1;   

	while( ch < '0' || ch > '9' ){if(ch=='-')sign=-1; ch=getcx();} 

	while(  ch >= '0' && ch <= '9' )    

	n = (n<<3)+(n<<1) + ch-'0', ch=getcx();

	n=n*sign;

}



int mod[MAX];

int first[2019];

vector<int> possible;

void pre() {

	memset(first, -1, sizeof first);

	int i;

	int x = 10;

	for(i=1;i<MAX;i++) {

		if(first[x] == -1) {

			first[x] = i;

		}

		x = (x*10) % 2019;

	}

	

	for(i=1;i<2019;i++) {

		if(first[i] != -1) possible.PB(i);

	}

}



vector<int> mods[2019];

int main() {

	string s;

	int i, j;

	pre();

	cin>>s;

	

	LL ans = 0;

	for(i=0;i<s.length();i++) {

		mod[i] = (i == 0) ? (s[i]-'0') % 2019 : (mod[i-1] * 10 + s[i] - '0') % 2019;

		if(mod[i] == 0) ans++;

	}

	

	int oc[2019][225] = {0};

	

	for(i=0;i<s.length();i++) {

		oc[mod[i]][i%224]++;

	}

	

	//cout<<possible.size()<<endl;

	

	for(i=0;i<s.length();i++) {

		oc[mod[i]][i%224]--;

		for(j=0;j<possible.size();j++) {

			int start = first[possible[j]];

			int pos = (i+start) % 224;

			int target = (mod[i] * possible[j]) % 2019;

			ans += oc[target][pos];

			

		} 

	}

	

	cout<<ans<<endl;

	

    return 0;

}




// Optimized solution to the same problem (faster):
/*

    Author : Nishant Gupta 2.0

*/

#include<bits/stdc++.h>



#define LL long long int

#define getcx getchar_unlocked

#define X first

#define Y second

#define PB push_back

#define MP make_pair

#define MAX 200005

#define MOD 1000000007

#define INF 0x3f3f3f3f

#define INFL (LL(1e18))

#define rep(i, a, n) for(i=a;i<n;i++)

#define rev(i, a, n) for(i=a;i>=n;i--)

#define MSV(A,a) memset(A, a, sizeof(A))

#define rep_itr(itr, c) for(itr = (c).begin(); itr != (c).end(); itr++)



using namespace std;



inline void inp( int &n ) { 

	n=0;    int ch=getcx();int sign=1;   

	while( ch < '0' || ch > '9' ){if(ch=='-')sign=-1; ch=getcx();} 

	while(  ch >= '0' && ch <= '9' )    

	n = (n<<3)+(n<<1) + ch-'0', ch=getcx();

	n=n*sign;

}



inline void inp1( LL &n ) { 

	n=0;    LL ch=getcx();LL sign=1;   

	while( ch < '0' || ch > '9' ){if(ch=='-')sign=-1; ch=getcx();} 

	while(  ch >= '0' && ch <= '9' )    

	n = (n<<3)+(n<<1) + ch-'0', ch=getcx();

	n=n*sign;

}

/*

int mod[MAX];

int first[2019];

vector<int> possible;

void pre() {

	memset(first, -1, sizeof first);

	int i;

	int x = 10;

	for(i=1;i<MAX;i++) {

		if(first[x] == -1) {

			first[x] = i;

		}

		x = (x*10) % 2019;

	}

	

	for(i=1;i<2019;i++) {

		if(first[i] != -1) possible.PB(i);

	}

}

*/

vector<int> mods[2019];

int main() {

	string s;

	int i, j;

	//pre();

	cin>>s;

	

	LL ans = 0;

	int r = 0;

	int x = 1;

	int freq[2019] = {0};

	for(i=s.length()-1;i>=0;i--) {

		r = (r + x*(s[i]-'0')) % 2019;

		x = (x*10)%2019;

		freq[r]++; 

	}

	

	

	for(i=1;i<2019;i++) {

		ans = ans + (freq[i]*(freq[i]-1))/2;

	}

	

	ans = ans + (freq[0]*(freq[0]+1))/2;

	

	cout<<ans<<endl;

	

    return 0;

}




// Fastest known solution for this problem: