Problem:
Score : 700 points

Problem Statement

Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order.

Iroha is looking for X,Y,Z-Haiku (defined below) in integer sequences.

Consider all integer sequences of length N whose elements are between 1 and 10, inclusive. Out of those 10^N sequences, how many contain an X,Y,Z-Haiku?

Here, an integer sequence a_0, a_1, ..., a_{N-1} is said to contain an X,Y,Z-Haiku if and only if there exist four indices x, y, z, w (0 ≦ x < y < z < w ≦ N) such that all of the following are satisfied:

a_x + a_{x+1} + ... + a_{y-1} = X

a_y + a_{y+1} + ... + a_{z-1} = Y

a_z + a_{z+1} + ... + a_{w-1} = Z

Since the answer can be extremely large, print the number modulo 10^9+7.

Constraints

3 ≦ N ≦ 40

1 ≦ X ≦ 5

1 ≦ Y ≦ 7

1 ≦ Z ≦ 5

Input

The input is given from Standard Input in the following format:

N X Y Z

Output

Print the number of the sequences that contain an X,Y,Z-Haiku, modulo 10^9+7.

Sample Input 1

3 5 7 5

Sample Output 1

1

Here, the only sequence that contains a 5,7,5-Haiku is [5, 7, 5].

Sample Input 2

4 5 7 5

Sample Output 2

34

Sample Input 3

37 4 2 3

Sample Output 3

863912418

Sample Input 4

40 5 7 5

Sample Output 4

562805100

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define rep(i,N) for(int i=0,i##_max=(N);i<i##_max;++i)
#define repp(i,l,r) for(int i=(l),i##_max=(r);i<i##_max;++i)
#define per(i,N) for(int i=(N)-1;i>=0;--i)
#define perr(i,l,r) for(int i=r-1,i##_min(l);i>=i##_min;--i)
#define all(arr) (arr).begin(), (arr).end()
#define SP << " " <<
#define SPF << " "
#define SPEEDUP cin.tie(0);ios::sync_with_stdio(false);
#define MAX_I INT_MAX //1e9
#define MIN_I INT_MIN //-1e9
#define MAX_UI UINT_MAX //1e9
#define MAX_LL LLONG_MAX //1e18
#define MIN_LL LLONG_MIN //-1e18
#define MAX_ULL ULLONG_MAX //1e19
typedef long long ll;
typedef pair<int,int> PII;
typedef pair<char,char> PCC;
typedef pair<ll,ll> PLL;
typedef pair<char,int> PCI;
typedef pair<int,char> PIC;
typedef pair<ll,int> PLI;
typedef pair<int,ll> PIL; 
typedef pair<ll,char> PLC; 
typedef pair<char,ll> PCL; 

inline void YesNo(bool b){ cout << (b?"Yes" : "No") << endl;}
inline void YESNO(bool b){ cout << (b?"YES" : "NO") << endl;}
inline void Yay(bool b){ cout << (b?"Yay!" : ":(") << endl;}

template<int MOD> struct Fp {
    ll val;
    constexpr Fp(ll v = 0) noexcept : val(v % MOD) {
        if (val < 0) v += MOD;
    }
    constexpr int getmod() { return MOD; }
    constexpr Fp operator - () const noexcept {
        return val ? MOD - val : 0;
    }
    constexpr Fp operator + (const Fp& r) const noexcept { return Fp(*this) += r; }
    constexpr Fp operator - (const Fp& r) const noexcept { return Fp(*this) -= r; }
    constexpr Fp operator * (const Fp& r) const noexcept { return Fp(*this) *= r; }
    constexpr Fp operator / (const Fp& r) const noexcept { return Fp(*this) /= r; }
    constexpr Fp& operator += (const Fp& r) noexcept {
        val += r.val;
        if (val >= MOD) val -= MOD;
        return *this;
    }
    constexpr Fp& operator -= (const Fp& r) noexcept {
        val -= r.val;
        if (val < 0) val += MOD;
        return *this;
    }
    constexpr Fp& operator *= (const Fp& r) noexcept {
        val = val * r.val % MOD;
        return *this;
    }
    constexpr Fp& operator /= (const Fp& r) noexcept {
        long long a = r.val, b = MOD, u = 1, v = 0;
        while (b) {
            long long t = a / b;
            a -= t * b; swap(a, b);
            u -= t * v; swap(u, v);
        }
        val = val * u % MOD;
        if (val < 0) val += MOD;
        return *this;
    }
    constexpr bool operator == (const Fp& r) const noexcept {
        return this->val == r.val;
    }
    constexpr bool operator != (const Fp& r) const noexcept {
        return this->val != r.val;
    }
    friend constexpr ostream& operator << (ostream &os, const Fp<MOD>& x) noexcept {
        return os << x.val;
    }
    friend constexpr istream& operator >> (istream &is, Fp<MOD>& x) noexcept {
        return is >> x.val;
    }
    friend constexpr Fp<MOD> modpow(const Fp<MOD> &a, long long n) noexcept {
        if (n == 0) return 1;
        auto t = modpow(a, n / 2);
        t = t * t;
        if (n & 1) t = t * a;
        return t;
    }
};


//const int MOD = 998244353;
//const int MOD = 1e6 + 3;
const int MOD = 1e9 + 7;
const int N_MAX = 2e5+5;
//mint:mod演算のための型
using mint = Fp<MOD>;
//型変換
#define MINT (mint)

//mod MOD上での階乗
mint fact[N_MAX];
mint finv[N_MAX];
void init(){
  fact[0] = fact[1] = MINT 1;
  for(int i = 2; i < N_MAX; i++){
    fact[i] = fact[i-1] * i;
  }
  finv[N_MAX-1] = MINT 1/fact[N_MAX-1];
  for(int i = N_MAX-1; i>0; --i){
    finv[i-1] = finv[i] * i;
  }
}

//mod MOD上での逆元
inline mint inv(mint n){
  return MINT 1/n;
}

//二項係数の計算(mod MOD上)
inline mint COM(int n, int k){
  if(n<k) return 0;
  if(k<0) return 0;
  return fact[n]*finv[k]*finv[n-k];
}

//順列計算
inline mint PER(int n, int k){
  if(n<k) return 0;
  if(k<0) return 0;
  return fact[n]*finv[n-k];
}

int main(void){
  SPEEDUP
  cout << setprecision(15);
  init();
  int N,X,Y,Z;cin >> N >> X >> Y >> Z;
  //方針：数値をx : 1<<(x-1)で保存する．
  //保存に必要なビット数(これ以前の情報は不要
  int n = (1<<(X+Y+Z))-1;
  //この集合を部分集合に含むならXYZができるのでダメ．
  int t = (1<<(Z-1)) | (1<<(Z+Y-1)) | (1<<(X+Y+Z-1));
  //i番目まで見て，jの状態からできる数
  mint dp[2][n+1];
  mint *pre,*nxt;
  pre = dp[0];
  nxt = dp[1];
  //0番目まで見て0の状態にできるのは1個(何もしない)
  pre[0] = 1;
  rep(i,N){
    rep(j,n+1){
      //i番目でjにできないなら意味がないので無視
      if(pre[j] == MINT 0) continue;
      repp(k,1,10+1){
        //次に作るパターン
        int r = (j<<k|(1<<(k-1)))&n;
        //パターンがXYZでないなら加算．
        if((r&t)!=t) nxt[r] += pre[j];
      }
    }
    swap(pre,nxt);
    rep(j,n+1) nxt[j] = 0;
  }
  swap(pre,nxt);
  mint ans = modpow(MINT 10,N);
  rep(i,n+1) ans -= nxt[i];
  cout << ans << endl;
  return 0;
}