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>

const int N(45), Mod(1e9 + 7);
int n, X, Y, Z, ans = 1, f[N][1 << 17];

int main()
{
	scanf("%d%d%d%d", &n, &X, &Y, &Z); int S = (1 << (X + Y + Z)) - 1;
	int H = (1 << (Z - 1)) | (1 << (Y + Z - 1)) | (1 << (X + Y + Z - 1));
	for (int i = 1; i <= n; i++) ans = 10ll * ans % Mod;
	for (int i = (f[0][0] = 1, 0); i < n; i++)
		for (int T = 0; T <= S; T++) if (f[i][T])
			for (int j = 1, s; j <= 10; j++)
				if (((s = ((T << j) | (1 << (j - 1))) & S) & H) != H)
					f[i + 1][s] = (f[i + 1][s] + f[i][T]) % Mod;
	for (int i = 0; i <= S; i++) ans = (ans - f[n][i] + Mod) % Mod;
	printf("%d\n", ans);
	return 0;
}
