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 PyPy3 (2.4.0) (judge verdict: Accepted)

# -*- coding: utf-8 -*-

import sys

def input(): return sys.stdin.readline().strip()
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e] * d for j in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(N=None): return list(MAP()) if N is None else [INT() for i in range(N)]
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
sys.setrecursionlimit(10 ** 9)
INF = 10 ** 18
MOD = 10 ** 9 + 7

N, X, Y, Z = MAP()
M = X + Y + Z

# XYZとの一致をチェックするビット列
check = 1<<(X+Y+Z-1) | 1<<(Y+Z-1) | 1<<(Z-1)
dp = list2d(N+1, 1<<(M-1), 0)
# dp[i][S] := i個目まで見て、前の値M-1以下までの集合がSの時の、XYZを含まないものの通り数
dp[0][0] = 1
msk1 = (1<<M) - 1
msk2 = (1<<(M-1)) - 1
for i in range(N):
    for S in range(1<<(M-1)):
        if not dp[i][S]:
            continue
        for j in range(10):
            # 1ビットずらして1を足してjビットずらしてマスクする
            S2 = (S<<1|1)<<j & msk1
            # 直前Mビットで使った値がXYZと一致していないか確認する
            if (S2 & check) != check:
                dp[i+1][S2&msk2] += dp[i][S]
                dp[i+1][S2&msk2] %= MOD
total = pow(10, N, MOD)
print((total-sum(dp[N])%MOD)%MOD)