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>
#include <stdlib.h>

int n,m,q;
int a[50],b[50],c[50],d[50];
int process();
int process_sub(int ps);
int ch();
int A[13];

int main()
{
    scanf("%d %d %d",&n,&m,&q);

    int i;

    for(i=0;i<q;i++)
        scanf("%d %d %d %d",&a[i],&b[i],&c[i],&d[i]);

    int max = process();
    printf("%d\n",max);

    return 0;
}

int process()
{
    int i;
    int ps = 0;
    int max=0,ans;

    for(i=1;i<=m;i++)
    {
        A[0] = i;
       ans =  process_sub(ps+1);
        if(max < ans)
            max = ans;
    }

    return max;
}

int process_sub(int ps)
{
    int max=0,ans;
    if(ps == n)
    {
        ans = ch();
        return ans;
    }

    int i;

    for(i = A[ps-1]; i<=m;i++)
    {
        A[ps] = i;
        ans = process_sub(ps+1);
        if(max < ans)
            max = ans;
    }

    return max;
}

int ch()
{
    int i;

    int sum_d = 0;

    for(i=0;i<q;i++)
    {
        if(A[b[i]-1] - A[a[i]-1] == c[i])
            sum_d += d[i];
    }

    

    return sum_d;
}