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<math.h>
int main()
{
    int n, m, q, i, c, k, score=0, max;
    scanf("%d%d%d", &n, &m, &q);

    int seq[n], ai[50], bi[50], ci[50], di[50];

    for(i=0; i<q; i++){
        scanf("%d%d%d%d",  &ai[i], &bi[i], &ci[i], &di[i]);
    }

    for(i=0; i<n; i++){
        seq[i]=1;
    }
    max=score;
    while(1){

        score=0;
        for(i=0; i<q; i++){
            if((seq[bi[i]-1]-seq[ai[i]-1])==ci[i]){
                score+=di[i];
            }
        }
        if(max<score){
            max=score;
        }

        if(seq[0]==m) break;

        seq[n-1]++;

        if(seq[n-1]>m){
            for(c=n-1; c>=0; c--){
                if(seq[c]>m){
                    seq[c-1]++;
                    for(k=c; k<n; k++){
                        seq[k]=seq[c-1];
                    }
                }
                else break;
            }
        }
    }

    printf("%d\n", max);
    return 0;
}