#include <iostream>
#include <sstream>
#include <iomanip>
#include <vector>
#include <string>
#include <map>
#include <set>
#include <algorithm>
#include <numeric>
#include <iterator>
#include <functional>
#include <complex>
#include <queue>
#include <stack>
#include <cmath>
#include <cassert>
#include <cstring>
using namespace std;
typedef long long LL;
typedef complex<double> CMP;

static const int MODVAL = 500500573;
struct mint
{
	int val;
	mint():val(0){}
	mint(int x):val(x%MODVAL) {}
	mint(LL  x):val(x%MODVAL) {}
};
mint operator+(mint x, mint y) { return x.val+y.val; }
mint operator-(mint x, mint y) { return x.val-y.val+MODVAL; }
mint operator*(mint x, mint y) { return LL(x.val)*y.val; }
mint POW(mint x, int e) {
	mint v = 1;
	for(;e;x=x*x,e>>=1)
		if(e&1)
			v=v*x;
	return v;
}
mint operator/(mint x, mint y) { return x * POW(y, MODVAL-2); }

class ProductAndSum { public:
	int getSum(int p2, int p3, int p5, int p7, int S) 
	{
		// Precomputing n!, 1/n!, and 111...111
		vector<mint> FACT(2501), FACT_INV(2501), ONES(2501);
		FACT[0] = FACT_INV[0] = 1;
		ONES[0] = 0;
		for(int n=1; n<FACT.size(); ++n)
		{
			FACT[n] = FACT[n-1] * n;
			FACT_INV[n] = 1 / FACT[n];
			ONES[n] = ONES[n-1]*10 + 1;
		}

		mint answer = 0;

		// Exhaustive search over the number of 4,6,8,9s : at most 33*50*50*100 < 8M iterations
		for(int v8=0; v8*3<=p2; ++v8)
		for(int v4=0; v4*2+v8*3<=p2; ++v4)
		for(int v9=0; v9*2<=p3; ++v9)
		for(int v6=0; v6+v4*2+v8*3<=p2 && v6+v9*2<=p3; ++v6)
		{
			// then, the number of 1,2,3s are computable
			int v[] = {-1, -1, p2-v8*3-v4*2-v6, p3-v9*2-v6, v4, p5, v6, p7, v8, v9};
			{
				int v1 = S;
				for(int i=2; i<=9; ++i)
					v1 -= i * v[i];
				if( v1 < 0 )
					continue;
				v[1] = v1;
			}
			int N = accumulate(v+1, v+10, 0);

			// Now, let's compute the sum of all integers, who have N digits and #d = v[d], in constant time!

			// It can be calculated as follows:
			//   Q: how many of the integers have, say "1" in the last digit?
			//   A: it is, of course, C(N-1, v[1]-1, v[2], ..., v[9])
			//      where C(k,a,b,..) is the # of ways of spliting k elements into groups of size a, b, ...
			// So, the sum of the least digits of the integers are
			//   X = \Sigma_{d=1..9} (d * C(N-1, v[1], v[2], ..., v[d]-1, ..., v[9]))
			// By the same argument, the sum of the second-last digits (10-no-kurai in japanese) is
			//   10 * X
			// For hundreds, the sum is 100*X, and so on. Eventualy, the sum of whole integers is
			//   t = (1+10+...+10^N) * X
			//     = 111...111 * \Sigma_{d=1..9} (d * C(N-1, v[1], v[2], ..., v[d]-1, ..., v[9]))
			// Let us simplify the formula by using C(k,a,b,...) = k! / a!b!...
			//   t = 111...111 * \Sigma_{d=1..9} (d * (N-1)! / v[1]! / ... / v[9]! * v[d])
			//     = 111...111 * (N-1)! / v[1]! / ... / v[9]! * \Sigma_{d=1..9} (d*v[d])
			//     = 111...111 * (N-1)! / v[1]! / ... / v[9]! * S
			// That's it!

			mint t = ONES[N] * FACT[N-1] * S;
			for(int i=1; i<=9; ++i)
				t = t * FACT_INV[v[i]];
			answer = answer + t;
		}
		return answer.val;
	}
};

// BEGIN CUT HERE
#include <ctime>
double start_time; string timer()
 { ostringstream os; os << " (" << int((clock()-start_time)/CLOCKS_PER_SEC*1000) << " msec)"; return os.str(); }
template<typename T> ostream& operator<<(ostream& os, const vector<T>& v)
 { os << "{ ";
   for(typename vector<T>::const_iterator it=v.begin(); it!=v.end(); ++it)
   os << '\"' << *it << '\"' << (it+1==v.end() ? "" : ", "); os << " }"; return os; }
void verify_case(const int& Expected, const int& Received) {
 bool ok = (Expected == Received);
 if(ok) cerr << "PASSED" << timer() << endl;  else { cerr << "FAILED" << timer() << endl;
 cerr << "\to: \"" << Expected << '\"' << endl << "\tx: \"" << Received << '\"' << endl; } }
#define CASE(N) {cerr << "Test Case #" << N << "..." << flush; start_time=clock();
#define END	 verify_case(_, ProductAndSum().getSum(p2, p3, p5, p7, S));}
int main(){

CASE(0)
	int p2 = 2; 
	int p3 = 0; 
	int p5 = 0; 
	int p7 = 0; 
	int S = 4; 
	int _ = 26; 
END
CASE(1)
	int p2 = 0; 
	int p3 = 0; 
	int p5 = 0; 
	int p7 = 0; 
	int S = 10; 
	int _ = 110109965; 
END
CASE(2)
	int p2 = 2; 
	int p3 = 0; 
	int p5 = 0; 
	int p7 = 0; 
	int S = 5; 
	int _ = 610; 
END
CASE(3)
	int p2 = 1; 
	int p3 = 1; 
	int p5 = 1; 
	int p7 = 1; 
	int S = 10; 
	int _ = 0; 
END
CASE(4)
	int p2 = 5; 
	int p3 = 5; 
	int p5 = 5; 
	int p7 = 5; 
	int S = 100; 
	int _ = 61610122; 
END
CASE(5)
	int p2 = 100; 
	int p3 = 100; 
	int p5 = 100; 
	int p7 = 100; 
	int S = 2500; 
	int _ = -1; 
END
/*
CASE(6)
	int p2 = ; 
	int p3 = ; 
	int p5 = ; 
	int p7 = ; 
	int S = ; 
	int _ = ; 
END
*/
}
// END CUT HERE
