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 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <string.h>
#define BMAX 16
#define MOD 1000000007

int main(void)
{
  int i, j, k, n, x, y, z, xyz, mask, idx = 0, res = 1, dp[2][1<<BMAX] = {0};
  scanf("%d %d %d %d", &n, &x, &y, &z);
  xyz = (1 << (x + y + z - 1)) | (1 << (y + z - 1)) | (1 << (z - 1));
  mask = ((1 << (x + y + z)) - 1) >> 1;
  dp[idx][0] = 1;
  for (i = 0; i < n; i++) {
    memset(dp[idx^1], 0, sizeof(dp[idx^1]));
    for (j = 0; j <= mask; j++) {
      if (dp[idx][j] > 0) {
        for (k = 0; k < 10; k++) {
          int used = (j << (k + 1)) | (1 << k);
          if ((used & xyz) != xyz) dp[idx^1][used&mask] = (dp[idx^1][used&mask] + dp[idx][j]) % MOD;
        }
      }
    }
    idx ^= 1;
  }
  for (i = 0; i < n; i++) res = (res * 10LL) % MOD;
  for (i = 0; i <= mask; i++) res = (res - dp[idx][i] + MOD) % MOD;
  printf("%d\n", res);
}