#pragma GCC optimization ("O3")

#include <bits/stdc++.h>

using namespace std;

using ll = long long;
using vec = vector<ll>;
using mat = vector<vec>;
using pll = pair<ll,ll>;

#define INF (1LL<<61)
#define MOD 1000000007LL 
//#define MOD 998244353LL
#define EPS (1e-10)

#define PR(x) cout << (x) << endl
#define PS(x) cout << (x) << " "
#define REP(i,m,n) for(ll (i)=(m),(i_len)=(n);(i)<(i_len);++(i))
#define FORE(i,v) for(auto (i):v)
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((ll)(x).size())
#define REV(x) reverse(ALL((x)))
#define ASC(x) sort(ALL((x)))
#define DESC(x) {ASC((x)); REV((x));}
#define BIT(s,i) (((s)>>(i))&1)
#define pb push_back
#define fi first
#define se second

template<class T> inline int chmin(T& a, T b) {if(a>b) {a=b; return 1;} return 0;}
template<class T> inline int chmax(T& a, T b) {if(a<b) {a=b; return 1;} return 0;}
class mint {
public:
    ll x;
    mint(ll x=0) : x((x%MOD+MOD)%MOD) {}
    mint operator-() const {return mint(-x);}
    mint& operator+=(const mint& a) {if((x+=a.x)>=MOD) x-=MOD; return *this;}
    mint& operator-=(const mint& a) {if((x+=MOD-a.x)>=MOD) x-=MOD; return *this;}
    mint& operator*=(const mint& a) {(x*=a.x)%=MOD; return *this;}
    mint operator+(const mint& a) const {mint b(*this); return b+=a;}
    mint operator-(const mint& a) const {mint b(*this); return b-=a;}
    mint operator*(const mint& a) const {mint b(*this); return b*=a;}
    mint pow(ll t) const {if(!t) return 1; mint a=pow(t>>1); return (t&1?*this*a:a)*a;}
    mint inv() const {return pow(MOD-2);}
    mint& operator/=(const mint& a) {return *this*=a.inv();}
    mint operator/(const mint& a) const {mint b(*this); return b/=a;}
};
istream &operator>>(istream& is, mint& a) {ll t; is>>t; a=t; return is;}
ostream &operator<<(ostream& os, const mint& a) {return os<<a.x;}
using mvec = vector<mint>;
using mmat = vector<mvec>;

mat matmul(mat A, mat B)
{
    ll N = SZ(A);
    mat C(N, vec(N, 0));
    REP(i,0,N) {
        REP(j,0,N) {
            REP(k,0,N) C[i][j] += A[i][k]*B[k][j], C[i][j] %= MOD;
        }
    }
    return C;
}

mat matpow(mat A, ll n)
{
    if(n == 0) {
        ll N = SZ(A);
        mat I(N, vec(N, 0));
        REP(i,0,N) I[i][i] = 1;
        return I;
    }
    mat T = matpow(A, n>>1);
    T = matmul(T, T);
    if(n&1) T = matmul(T, A);
    return T; 
}

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

int main()
{
    ll N, P, C;
    cin >> N >> P >> C;

    vec A = {2, 3, 5, 7, 11, 13};
    vec B = {4, 6, 8, 9, 10, 12};
    dp[0][0][0] = 1;
    dp[1][0][0] = 1;
    REP(x,0,6) REP(y,0,P) REP(i,0,y*13+1) if(dp[0][y][i]) (dp[0][y+1][i+A[x]] += dp[0][y][i]) %= MOD;
    REP(x,0,6) REP(y,0,C) REP(i,0,y*12+1) if(dp[1][y][i]) (dp[1][y+1][i+B[x]] += dp[1][y][i]) %= MOD;
    REP(x,0,651) REP(y,0,601) (single[x+y] += dp[0][P][x]*dp[1][C][y]) %= MOD;

    ll M = P*13+C*12;
    mat X(M, vec(M));
    REP(i,0,M-1) X[i][i+1] = 1;
    REP(i,1,M+1) X[M-1][M-i] = single[i];
    X = matpow(X, N+M-1);

    ll ans = 0;
    REP(i,0,M) ans += X[0][i];
    PR(ans%MOD);

    return 0;
}

/*

13*5+12*5=125

*/