Problem:
Score : 300 points

Problem Statement

Given are positive integers N, M, Q, and Q quadruples of integers ( a_i , b_i , c_i , d_i ).

Consider a sequence A satisfying the following conditions:

A is a sequence of N positive integers.

1 \leq A_1 \leq A_2 \le \cdots \leq A_N \leq M.

Let us define a score of this sequence as follows:

The score is the sum of d_i over all indices i such that A_{b_i} - A_{a_i} = c_i. (If there is no such i, the score is 0.)

Find the maximum possible score of A.

Constraints

All values in input are integers.

2 ≤ N ≤ 10

1 \leq M \leq 10

1 \leq Q \leq 50

1 \leq a_i < b_i \leq N ( i = 1, 2, ..., Q )

0 \leq c_i \leq M - 1 ( i = 1, 2, ..., Q )

(a_i, b_i, c_i) \neq (a_j, b_j, c_j) (where i \neq j)

1 \leq d_i \leq 10^5 ( i = 1, 2, ..., Q )

Input

Input is given from Standard Input in the following format:

N M Q
a_1 b_1 c_1 d_1
:
a_Q b_Q c_Q d_Q

Output

Print the maximum possible score of A.

Sample Input 1

3 4 3
1 3 3 100
1 2 2 10
2 3 2 10

Sample Output 1

110

When A = \{1, 3, 4\}, its score is 110. Under these conditions, no sequence has a score greater than 110, so the answer is 110.

Sample Input 2

4 6 10
2 4 1 86568
1 4 0 90629
2 3 0 90310
3 4 1 29211
3 4 3 78537
3 4 2 8580
1 2 1 96263
1 4 2 2156
1 2 0 94325
1 4 3 94328

Sample Output 2

357500

Sample Input 3

10 10 1
1 10 9 1

Sample Output 3

1

Solution in Java (OpenJDK 11.0.6) (judge verdict: Runtime Error)

import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {

		Scanner scan = new Scanner(System.in);

		int N = scan.nextInt();
		int M = scan.nextInt();
		int Q = scan.nextInt();
		List<int[]> list = new ArrayList<>();
		long answer = 0;

		for (int i = 0; i < Q; i++) {
			int[] temp = new int[4];
			for (int k = 0; k < 4; k++) {
				temp[k] = scan.nextInt();
			}
			list.add(temp);
		}

		int[] d = new int[M + 1];
		for (int i = 1; i <= M; i++) d[i] = 1;

		int[] A = new int[M];
		while (d[0] <= 0){
			long temp = 0;
			for (int i = 1; i <= N; i++) A[i - 1] = d[i];

			for (int[] nums : list) {
				if (A[nums[1] - 1] - A[nums[0] - 1] == nums[2]) temp += nums[3];
			}
			answer = Math.max(temp, answer);

			for (int i = N; 0 <= i; i--){
				d[i]++;
				for (int k = i + 1; k <= N; k++) d[k] = d[k - 1];
				if (d[i] <= M)break;
			}
		}

		System.out.println(answer);
	}
}