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 C (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

#include <math.h>
#include <stdio.h>
int main() {
	int n, m, q, A[10], a[50], b[50], c[50], d[50];
	int sum, max = 0;
	scanf("%d%d%d", &n, &m, &q);
	for (int i = 0; i < q; i++) {
		scanf("%d%d%d%d", &a[i], &b[i], &c[i], &d[i]);
	}
	for (int fi = 1; fi <= m; fi++) {
		int l = pow(m, n - 1);
		A[0] = fi;
		for (int bit = 0; bit < l; bit++) {
			int num = bit;
			for (int i = 1; i < n; i++) {
				A[i] = A[i - 1] + (num % m);
				num /= m;
			}
			if (A[n - 1] > m) {
				;
			} else {
				/*
				for (int i = 0; i < n; i++) {
					printf("%d ", A[i]);
				}
				printf("\n");
				*/
				sum = 0;
				for (int i = 0; i < q; i++) {
					if (A[b[i] - 1] - A[a[i] - 1] == c[i]) {
						sum += d[i];
					}
				}
				if (sum > max) {
					max = sum;
				}
			}
		}
	}
	printf("%d\n", max);
	return 0;
}