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.

Examples

Input

3 5 7 5


Output

1


Input

4 5 7 5


Output

34


Input

37 4 2 3


Output

863912418


Input

40 5 7 5


Output

562805100
#include <bits/stdc++.h>
#define ll long long
using namespace std;
const ll N=42, Max=(1<<17)+10, MOD=1e9+7;
ll n, x, y, z, f[N][Max], sum=1, ans;
inline ll M(ll a){return a%MOD;}

void work(){
	for(int i = 0; i < n; i++) sum = M(sum*10);
	ll cnt = (1<<x+y+z) -1;
	ll tot = (1<<x-1) | (1<<x+y-1) | (1<<x+y+z-1); 
	f[0][0] = 1;
	for(int i = 1; i <= n; i++) 
		for(int j = 0; j <= cnt; j++) 
			for(int h = 1;h <= 10; h++){
				ans = (j<<h)|(1<<h-1);
				ans &= cnt;
				if((ans&tot)!=tot) f[i][ans] = M(f[i][ans]+f[i-1][j]);
			}
	for(int i = 0; i <= cnt; i++)
		sum = M(sum-f[n][i]+MOD);
}

int main(){
	scanf("%lld%lld%lld%lld",&n,&x,&y,&z);
	work();
	printf("%lld",sum);
	return 0;
}