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: Wrong Answer)

#include <stdio.h>

int main(void) {
 int d[3];
 scanf("%d %d %d", &d[0],&d[1],&d[2]);

 int A[d[2]];
  int B[d[2]];
   int C[d[2]];
    int D[d[2]];
     int a[d[0]];
    int scorem=0;
   for (int j = 0; j < d[0]; j++)

            {
                a[j] = 1;

            }
         a[d[0]-1] = a[d[0] - 1]-1;
      for (int i = 0; i < d[2]; i++)
            {
                int tmp[4];
                scanf("%d %d %d %d",&tmp[0],&tmp[1],&tmp[2],&tmp[3]);
                A[i] = tmp[0];
                B[i] = tmp[1];
                C[i] = tmp[2];
                D[i] = tmp[3];
            }
        
      while (1)
            {
                if (a[0] == d[1]) break;
                int p_ = d[0]-1;
               for(int i = d[0]-1;i >=1;i--)
                {
                    if (a[i] - a[i - 1] >= 1) p_ = i-1;
                }
                //reset
                for (int i = 0; i <p_; i++)
                {
                    a[i] = 1;
                }
                a[p_]++;

                //得点
                int score = 0;
                for (int i = 0; i < d[2]; i++)
                {
                    int c = a[B[i] -1] - a[A[i] - 1]; //a_Bi - a_Ai
                    int c_ = C[i];//Ci
                    if (c == c_)
                    {
                        score += D[i];
                     //  RESULT(score);
                    }
                }
                if (scorem < score) scorem = score;

            }
            printf("%d",scorem);

  return 0;
}