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 (Clang 10.0.0) (judge verdict: Wrong Answer)

#include <inttypes.h>
#include <math.h>
#include <stdbool.h>
#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main(void)
{
    intmax_t N, M, Q, a[50], b[50], c[50], d[50];
    intmax_t A[10];
    intmax_t ans = 0;

    scanf("%jd%jd%jd", &N, &M, &Q);
    for (size_t i = 0; i < Q; i++) {
        scanf("%jd%jd%jd%jd", &a[i], &b[i], &c[i], &d[i]);
    }

    for (A[0] = 0; A[0] <= M; A[0]++) {
        for (A[1] = A[0]; A[1] <= M; A[1]++) {
            for (A[2] = A[1]; A[2] <= M; A[2]++) {
                for (A[3] = A[2]; A[3] <= M; A[3]++) {
                    for (A[4] = A[3]; A[4] <= M; A[4]++) {
                        for (A[5] = A[4]; A[5] <= M; A[5]++) {
                            for (A[6] = A[5]; A[6] <= M; A[6]++) {
                                for (A[7] = A[6]; A[7] <= M; A[7]++) {
                                    for (A[8] = A[7]; A[8] <= M; A[8]++) {
                                        for (A[9] = A[8]; A[9] <= M; A[9]++) {
                                            intmax_t tmp = 0;
                                            for (size_t i = 0; i < Q; i++) {
                                                if (A[b[i] - 1] - A[a[i] - 1] == c[i]) {
                                                    tmp += d[i];
                                                }
                                            }
                                            if (tmp > ans)
                                                ans = tmp;
                                        }
                                    }
                                }
                            }
                        }
                    }
                }
            }
        }
    }

    printf("%jd\n", ans);

    return 0;
}