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 <algorithm>
#include <bitset>
#include <cassert>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <deque>
#include <functional>
#include <iomanip>
#include <iostream>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <utility>
#include <vector>

#define REP(i,s,n) for(int i=(int)(s);i<(int)(n);i++)

using namespace std;
typedef long long int ll;
typedef vector<int> VI;
typedef pair<int, int> PI;
const double EPS=1e-9;


const ll mod = 1e9 + 7;

const int N = 41;
const int S = 1 << 18;
ll dp[N][S];
int next_tbl[S][11];
int n, x, y, z;
int qs;

int next_state_nfa(int q, int s) {
  int lim;
  assert (q < qs);
  if (q < x) {
    lim = x;
  } else if (q < x + y) {
    lim = x + y;
  } else if (q < x + y + z) {
    lim = x + y + z;
  } else {
    return x + y + z;
  }
  
  if (q + s <= lim) {
    return q + s;
  }
  return -1;
}

int next_state(int q, int s) {
  int bits = 1;
  REP(i, 0, qs) {
    if (q & (1 << i)) {
      int r = next_state_nfa(i, s);
      if (r >= 0) bits |= 1 << r;
    }
  }
  return bits;
}

int main(void){
  cin >> n >> x >> y >> z;
  qs = x + y + z + 1;
  dp[0][1] = 1;
  REP(bits, 0, 1 << qs) {
    REP(s, 1, 11) {
      next_tbl[bits][s] = next_state(bits, s);
    }
  }
  REP(i, 1, n + 1) {
    REP(bits, 0, 1 << qs) {
      REP(s, 1, 11) {
	int next = next_tbl[bits][s];
	dp[i][next] += dp[i - 1][bits];
	dp[i][next] %= mod;
      }
    }
  }
  ll sum = 0;
  REP(bits, 1 << (qs - 1), 1 << qs) { // final states
    sum += dp[n][bits];
    sum %= mod;
  }
  cout << sum << endl;
}