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: Accepted)

#include <stdio.h>
#include <string.h>

int next_seq(int *seq, const int len, const int m)
{
    int carry = 0, last, last_value;
    for (int i = len - 1; i >= 0; i--)
    {
        if (seq[i] >= m && i > 0)
        {
            seq[i] = seq[i - 1] + 1;
        }
        else if (seq[i] >= m)
        {
            carry = 1;
        }
        else
        {
            seq[i]++;
            last = i;
            last_value = seq[i];
            break;
        }
    }

    for (int i = last + 1; i < len; i++)
    {
        seq[i] = last_value;
    }

    return carry;
}

int main()
{
    int n, m, q, max = 0;
    int a[50] = {
        0,
    };
    int b[50] = {
        0,
    };
    int c[50] = {
        0,
    };
    int d[50] = {
        0,
    };

    int seq[11] = {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1};

    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]);
    }

    seq[n - 1] = 0;

    while (!next_seq(seq, n, m))
    {
        int sum = 0;

        for (int i = 1; i <= q; i++)
        {
            if (seq[b[i - 1] - 1] - seq[a[i - 1] - 1] == c[i - 1])
                sum += d[i - 1];
        }
        /*
        for (int i = 0; i < n; i++)
        {
            printf("%02d ", seq[i]);
        }
        */
        max = max < sum ? sum : max;
        //printf(" %d\n", sum);
    }

    printf("%d", max);

    return 0;
}