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 <string.h>
#include <stdio.h>
#include <stdlib.h>
#include<math.h>
#define ll long long
int main(){
  ll int n,m,q,ans=0,max=0;
  scanf("%lld%lld%lld",&n,&m,&q);
  ll int a[q],b[q],c[q],d[q],A[11];
  for(int i=1;i<=q;i++){
    scanf("%lld%lld%lld%lld",&a[i],&b[i],&c[i],&d[i]);
  }
  for(int i=1;i<=m;i++){
    A[1]=i;
    for(int j=i;j<=m;j++){
      A[2]=j;
      for(int k=j;k<=m;k++){
        A[3]=k;
        for(int l=k;l<=m;l++){
          A[4]=l;
          for(int z=l;z<=m;z++){
            A[5]=z;
            for(int y=z;y<=m;y++){
              A[6]=y;
              for(int o=y;o<=m;o++){
                A[7]=o;
                for(int p=o;p<=m;p++){
                  A[8]=p;
                  for(int r=p;r<=m;r++){
                    A[9]=r;
                    for(int s=r;s<=m;s++){
                      A[10]=s;
  for(int u=1;u<=q;u++){
    if(A[b[u]]-A[a[u]]==c[u]){
      ans+=d[u];
    }
  }
                      if(max<ans){
                        max=ans;
                      }
                      ans=0;
                    }
                  }
                }
              }
            }
          }
        }
      }
    }
  }
  printf("%lld",max);
  return 0;
}