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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
#define rep(i,n)for(int i=0;i<(n);i++)
#define MOD 1000000007
using namespace std;
typedef long long ll;

int dp[50][1 << 16], mem[1 << 16][11];
int n, x, y, z;
int ok(int v) {
	bool a = false, b = false;
	int c = 0, d = 0;
	for (int i = 0; i < 17; i++) {
		c++;
		if (v >> i & 1) {
			if (!a&&c == z)a = true, c = 0;
			if (a && !b&&c == y)b = true, c = 0;
			if (a&&b&&c == x)return -1;
		}
		if (i != 16)d |= (v >> i & 1) << i;
	}
	return d;
}
int main() {
	scanf("%d%d%d%d", &n, &x, &y, &z);
	rep(i, 1 << 16) {
		for (int j = 1; j <= 10; j++) {
			int t = i << j; t |= 1 << (j - 1);
			mem[i][j] = ok(t);
		}
	}
	dp[0][0] = 1;
	rep(i, n)rep(j, 1 << 16) {
		if (dp[i][j] == 0)continue;
		for (int k = 1; k <= 10; k++) {
			if (mem[j][k] != -1)(dp[i + 1][mem[j][k]] += dp[i][j]) %= MOD;
		}
	}
	ll ans = 0, b = 1;
	rep(i, n)(b *= 10) %= MOD;
	rep(j, 1 << 16)(ans += dp[n][j]) %= MOD;
	printf("%lld\n", (b - ans + MOD) % MOD);
}