#include <bits/stdc++.h>
using namespace std;
typedef signed long long ll;

#undef _P
#define _P(...) (void)printf(__VA_ARGS__)
#define FOR(x,to) for(x=0;x<to;x++)
#define FORR(x,arr) for(auto& x:arr)
#define ITR(x,c) for(__typeof(c.begin()) x=c.begin();x!=c.end();x++)
#define ALL(a) (a.begin()),(a.end())
#define ZERO(a) memset(a,0,sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
//-------------------------------------------------------

ll N;
int P,C,M;
int A[6]={2,3,5,7,11,13};
int B[6]={4,6,8,9,10,12};
ll mo=1000000007;

ll dp[2][400][4000];
ll dp2[8000];
ll single[8010];

vector<ll> mult(vector<ll>& v,vector<ll>& v2) {
	int i,j;
	int M=v.size();
	vector<ll> t(2*M,0);
	FOR(i,M) FOR(j,M) {
		t[i+j] += v[i]*v2[j];
		if(t[i+j]>=mo) t[i+j]%=mo;
	}
	for(i=2*M-2;i>=M;i--) if(t[i]) {
		t[i]%=mo;
		for(j=1;j<=M;j++) t[i-j] += single[j]*t[i];
	}
	FOR(i,M) t[i]%=mo;
	t.resize(M);
	return t;
}

void solve() {
	int i,j,k,l,r,x,y; string s;
	
	cin>>N>>P>>C;
	dp[0][0][0]=dp[1][0][0]=1;
	
	FOR(x,6) FOR(y,P) FOR(i,y*13+1) if(dp[0][y][i]) (dp[0][y+1][i+A[x]] += dp[0][y][i])%=mo;
	FOR(x,6) FOR(y,C) FOR(i,y*12+1) if(dp[1][y][i]) (dp[1][y+1][i+B[x]] += dp[1][y][i])%=mo;
	FOR(x,651) FOR(y,601) (single[x+y] += dp[0][P][x]*dp[1][C][y])%=mo;
	M=P*13+C*12;
	dp2[0]=1;
	
	for(i=1;i<=2*M+2;i++) {
		FOR(x,M+1) if(i>=x) dp2[i] += dp2[i-x]*single[x]%mo;
		dp2[i] %= mo;
	}
	
	ll tot=0;
	if(N<=2*M) {
		for(ll v=max(0LL,N-M);v<N;v++) {
			for(x=1;x<=M;x++) if(v+x>=N) tot += dp2[v]*single[x]%mo;
			tot %= mo;
		}
	}
	else {
		
		ll vv=N-M;
		vector<ll> R(M,0),V(M,0);
		R[0]=V[1]=1;
		while(vv) {
			if(vv%2) R=mult(R,V);
			V=mult(V,V);
			vv/=2;
		}
		
		for(ll v=max(0LL,N-M);v<N;v++) {
			ll pat=0;
			for(x=0;x<=M-1;x++) pat += R[x]*dp2[v-(N-M)+x] % mo;
			pat %= mo;
			for(x=1;x<=M;x++) if(v+x>=N) tot += pat*single[x]%mo;
			tot %= mo;
		}
	}
	
	cout<<tot%mo<<endl;
}


int main(int argc,char** argv){
	string s;int i;
	if(argc==1) ios::sync_with_stdio(false);
	FOR(i,argc-1) s+=argv[i+1],s+='\n';
	FOR(i,s.size()) ungetc(s[s.size()-1-i],stdin);
	solve(); return 0;
}