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++ (GCC 9.2.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
#define title "title"
#define ll long long
#define ull unsigned ll
#define fix(x) fixed<<setprecision(x)
#define pii pair<int,int>
#define vint vector<int>
#define pb push_back
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define red(i,a,b) for(int i=(a);i>=(b);i--)
#define db double
#define ld long db
using namespace std;
void Freopen(){
	freopen(title".in","r",stdin);
	freopen(title".out","w",stdout);
}
int read(){
	int g=0,f=1;
	char ch=getchar();
	while(ch<'0'||'9'<ch){if(ch=='-')f=-1;ch=getchar();}
	while('0'<=ch&&ch<='9'){g=g*10+ch-'0';ch=getchar();}
	return g*f;
}
const int mod=1e9+7;
int n,x,y,z,f[45][1<<17][2],status;
signed main(){
	//freopen("a","r",stdin);
	n=read(),x=read(),y=read(),z=read();
	status=(1<<x+y+z-1)+(1<<y+z-1)+(1<<z-1);
	f[0][0][0]=1;
	rep(i,0,n-1)rep(s,0,(1<<x+y+z)-1)rep(opt,0,1)if(f[i][s][opt]){
		rep(j,1,10){
			int sta=((s<<j)+(1<<(j-1)))&((1<<x+y+z)-1),top=opt;
			if((sta&status)==status)top=1;
			f[i+1][sta][top]=(f[i+1][sta][top]+f[i][s][opt])%mod;
		}
	}
	int ans=0;rep(i,0,(1<<x+y+z)-1)ans=(ans+f[n][i][1])%mod;return cout<<ans,signed();
}