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>
#define ull unsigned long long int
#define ll long long
#define rep(x) for(A[x]=A[x-1];A[x]<=10;A[x]++)

ll max(ll a,ll b){
  if(b>a)return b;
  return a;
};

typedef struct quatt{
  int a,b,c;
  long d;
}quat;

ll hyouka(quat q[],int n,int A[]){
  ll sum=0;
  for(int k=0;k<n;k++)
    if(A[q[k].b]-A[q[k].a]==q[k].c)
      sum+=q[k].d;
  return sum;
};
  


int main(){
  int N,M,Q;
  int A[11]={1,1,1,1,1,1,1,1,1,1,1};
  ll ans=0;
  scanf("%d%d%d",&N,&M,&Q);
  quat q[Q];
  for(int j=0;j<Q;j++){
    scanf("%d%d%d%ld",&q[j].a,&q[j].b,&q[j].c,&q[j].d);
  };
  for(A[1]=1;A[1]<=10;A[1]++)
    rep(2)rep(3)rep(4)rep(5)rep(7)rep(8)rep(9)rep(10){
    ans=max(hyouka(q,Q,A),ans);
  };
  printf("%lld",ans);      
};