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 e,f, i, j, k, l, m, n, o, p, q, r, s, u, v, a[50], b[50], c[50], d[50], t[11],count=0,max=0;
  scanf("%d%d%d",&n, &m, &q);
  for(e=0;e<q;e++){
    scanf("%d%d%d%d", &a[e], &b[e], &c[e], &d[e]);
  }
  for(i=1;i<=m;i++){
    t[1]=i;
    for(j=i;j<=m;j++){
      t[2]=j;
      for(k=j;k<=m;k++){
        t[3]=k;
        for(l=k;l<=m;l++){
          t[4]=l;
          for(o=l;o<=m;o++){
            t[5]=o;
            for(p=o;p<=m;p++){
              t[6]=p;
              for(r=o;r<=m;r++){
                t[7]=r;
                for(s=r;s<=m;s++){
                  t[8]=s;
                  for(u=s;u<=m;u++){
                    t[9]=u;
                    for(v=u;v<=m;v++){
                      t[10]=v;
                      for(f=0;f<q;f++){
                        if(t[b[f]]-t[a[f]]==c[f]){
                          count+=d[f];
                        }
                      }
                      if(count>max){
                        max=count;
                      }
                      count=0;
                    }
                  }
                }
              }
            }
          }
        }
      }
    }
  }
  printf("%d",max);      
  return 0;
}