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 <cstdio>
#define ll long long
const int p = 1e9+7;
int n, X, Y, Z;
int f[42][1<<17];
int main() {
	scanf("%d%d%d%d", &n, &X, &Y, &Z);
	f[0][0] = 1;
	int Haiku = (1<<(Z-1)) | (1<<(Y+Z-1)) | (1<<(X+Y+Z-1));
	int mask = (1<<(X+Y+Z))-1, ans = 1;
	for(int i = 1; i <= n; i ++ ) ans = ans*10ll%p;
	for(int i = 0; i < n; i ++ ) for(int S = 0; S <= mask; S ++ ) if(f[i][S]) {
		for(int j = 1; j <= 10; j ++ ) {
			int to = ((S<<j) | (1<<(j-1))) & mask;
			if((to&Haiku) != Haiku) f[i+1][to] = (f[i+1][to]+f[i][S])%p;
		}
	}
	for(int i = 0; i <= mask; i ++ ) ans = (ans-f[n][i]+p)%p;
	printf("%d\n", ans);
}
