#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;
vector<int> pattern;
class Kitamasa{
   private:
      vector<int> C; // a[n+1] = C[0] a[n] + C[1] a[n-1]+...C[S-1] a[n-S+1] (n=0,1,...)
      int mod;
   public:
      Kitamasa(vector<int> p){
         C=p;mod=1000000007;
      }
      vector<int> mul(vector<int> u, vector<int> v){
         // a[x] = (u[0] a[S-1] + u[1] a[S-2] +, ... + u[S-1] a[0])
         // a[y] = (v[0] a[S-1] + v[1] a[S-2] +, ... + v[S-1] a[0])
         // our purpose is to calculate w such that
         // a[x+y] = (w[0] a[S-1] + w[1] a[S-2] +, ... + w[S-1] a[0])  
         // a[x+y] = (u[0]a[y+S-1]+u[1]a[y+S-2]+...+u[S-1]a[y])     
         //        = (u[0](v[0]a[2S-2]+v[1]a[2S-3]+...+v[S-1]a[S-1])
         //          +u[1](v[0]a[2S-3]+v[1]a[2S-4]+...+v[S-1]a[S-2])
         //          +...)
         //        = sum(u[i]v[j]a[k]) where 0<=i,j<=S-1, i+j+k=2S-2
         int S=u.size();
         vector<int> ret(2*S-1,0);
         for( int i = 0 ; i < S; i++){
            for( int j=0;j<S;j++){
               ret[2*S-2-i-j]=(1LL*ret[2*S-2-i-j]+u[i]*1LL*v[j])%mod;
            }
         }
         // now a[x+y]=ret[2S-2] a[2S-2]+ret[2S-3] a[2S-3]+...ret[0] a[0].
         // be carefule that order of ret is inverse of order of w

         // move ret[S+i] into ret[S+i-1]... (i=0,1,2,...,S-1)
         // the key concept is, ret[S+i]a[S+i] = ret[S+i](C[0]a[S+i-1]+C[1]a[S+i-2]+....C[S-1]a[i])
         for( int i = 2*S-2; i>=S; i--){
            for( int j = 0 ; j < S; j++){
               ret[i-1-j]=(1LL*ret[i-1-j]+ret[i]*1LL*C[j])%mod;
            }
         }
         ret.resize(S);
         reverse(ret.begin(),ret.end());
         return ret;
      }
};

int main(){
   long long N;
   int P,C;
   cin>>N>>P>>C;
   int T=P*13+C*12+1;
   int dp[T][P+1][6];
   int dp2[T][C+1][6];
   int pdice[]={2,3,5,7,11,13};
   int cdice[]={4,6,8,9,10,12};
   memset(dp,0,sizeof(dp));
   memset(dp2,0,sizeof(dp2));
   dp[0][0][0]=1;
   for( int i=1 ; i<=P; i++ )for( int j=0;j<T;j++) for( int k=0; k<6; k++ ){//fill dp[j][i][k]
      for( int l=0;l<=k; l++){
         if(j-pdice[k]>=0)dp[j][i][k]+=dp[j-pdice[k]][i-1][l];
      }
   }
   for(int i=0;i<T;i++){
      for(int k=0; k<6; k++){
         dp2[i][0][0]+=dp[i][P][k];
      }
   }
   for( int i=1 ; i<=C; i++ )for( int j=0;j<T;j++) for( int k=0; k<6; k++ ){//fill dp2[j][i][k]
      for( int l=0;l<=k; l++){
         if(j-cdice[k]>=0)dp2[j][i][k]+=dp2[j-cdice[k]][i-1][l];
      }
   }
   pattern=vector<int>(T,0);
   for(int i=0;i<T;i++){
      for( int k=0;k<6;k++){
         pattern[i]+=dp2[i][C][k];
      }
   }
   int S=T-1;
   // a[n+1] = p[0] a[n] + p[1] a[n-1]+...p[S-1] a[n-S+1] (n>=0,p[i]=pattern[i+1])
   // initial value: a[0]=1,a[-1]=1,a[-2]=1,.... # to regard a[n] as num of the way to over n 
   // a[1] = p[0]a[0]+ p[1]a[-1]+....p[S-1]a[-S+1]  
   // let's denote coefficients as v[1]=(p[0],...,p[S-1])
   // a[2] = p[0]a[1]+ p[1]a[0]+....p[S-1]a[-S+2]
   //      = (p[1]+p[0]^2)a[0]+(p[2]+p[1]p[0])a[1]...(p[S-1]+p[S-2]p[0])a[-S+2]+p[S-1]a[-S+1]
   // this can be described as v[2]=(p[1]+p[0]^2,...,p[-S+2]+p[-S+1]p[0],p[S-1])
   // Now, let's calculate v[N] with initial value is a[0]=a[-1]=a[-2]=...=a[-S+1]=1
   // this can be regarded as calculate V[N+S-1] with initial value A[S-1]=...=A[0]=1
   // note that V[0]=v[-S+1]=(0,0,0,...,1), V[1]=(0,0,0,...,1,0);
   int EXP=N+S-1;
   vector<int> v(S,0);
   for( int i =0 ; i < S; i++ ){
      v[i]=pattern[i+1];
      cout << v[i] <<", ";
   }
   cout << endl;
   long long ans = 0;
   int mod=1000000007;
      Kitamasa ktms(v);
      vector<int> a(S,0);
      vector<int> b(S,0);
      a[S-1]=1;
      b[S-2]=1;
      cout << EXP << endl;
      int off=1;
      int cnt=0;
      while( EXP > 0 ){
         if( EXP%2==1 ){
            cnt+=off;
            a=ktms.mul(a,b);
            /*
            cout << cnt<<": ";
            for( int i =0 ; i < S; i++ ){ cout << a[i]<<"+";}
            cout << endl;
         */
         }
         off*=2;
         b=ktms.mul(b,b);
         /*
         cout << off<<": ";
         for( int i =0 ; i < S; i++ ){ cout << b[i]<<"+";}
         cout << endl;
         */
         EXP/=2;
      }
      for( int i = 0 ; i <S; i++ ){
         ans+=a[i];
         ans%=mod;
      }
   cout << ans << endl;
   return 0;
}