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 <cstdio>
#include <cstring>
#include <iostream>
using namespace std;

#define ri register int
#define mod 1000000007

int bit[60];
int N, x, y, z;
int f[43][140005];

int main() {
    cin >> N >> x >> y >> z;
    for(ri i = 0; i <= 25; i ++) bit[i] = 1 << i;
    
    int ans = 1;
    for(ri i = 1; i <= N; i ++) 
    ans = 1ll * ans * 10 % mod;
    
    f[0][0] = 1;
    int gg = bit[z - 1] | bit[y + z - 1] | bit[x + y + z - 1];
    int lim = bit[x + y + z] - 1;

    for(ri i = 1; i <= N; i ++)
    for(ri S = 0; S <= lim; S ++)
    for(ri v = 1; v <= 10; v ++) {
        int T = (S << v) | bit[v - 1]; T &= lim;
        if((T & gg) != gg) f[i][T] = (f[i][T] + f[i - 1][S]) % mod;
    }
    
    for(ri S = 0; S <= lim; S ++)
    if((S & gg) != gg) ans = (ans - f[N][S] + mod) % mod;
    printf("%d\n", ans);
    return 0;
}