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<algorithm>
using namespace std;

typedef long long ll;
const ll MAXM=131072;
const ll MOD=1e9+7;

ll ksm(ll x,ll y){
    ll cnt=1;
    while(y){
        if(y&1) cnt=(1ll*cnt*x)%MOD;
        x=(1ll*x*x)%MOD;
        y>>=1;
    }
    return cnt;
}

ll n,x,y,z;
ll d[50][MAXM+100];

bool check(ll now){
    if((now&(1<<(x+y+z-1)))&&(now&(1<<(y+z-1)))&&(now&(1<<(z-1))))
        return 1;
    return 0;
}

int main(){
    scanf("%lld%lld%lld%lld",&n,&x,&y,&z);
    d[0][0]=1ll;
    ll ans=0ll;
    for(ll i=0;i<n;i++)
        for(ll j=0;j<MAXM;j++){
            if(!d[i][j]) continue;
            //prllf("%d %d\n",i,j);
            for(ll k=1;k<=10;k++){
                ll v=((j<<k)+(1<<(k-1)))%MAXM;
                if(check(v)){
                    ans+=d[i][j]*ksm(10,n-i-1);
                    ans%=MOD;
                }
                else{
                    d[i+1][v]+=d[i][j];
                    d[i+1][v]%=MOD;
                }
            }
        }
    printf("%lld\n",ans);
}