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()
{
    int N=0,M=0,Q=0,abcd[200]={0},A[10]={0},sum=0,newsum=0,a1,a2,a3,a4,a5,a6,a7,a8,a9,a10;
    scanf("%d %d %d",&N,&M,&Q);
    for(int j=0;j<4*Q;j++)
	scanf("%d",&abcd[j]);

    
    for(a1=1;a1<=M;a1++){
	A[0]=a1;
	for(a2=1;a2<=M;a2++){
	    A[1]=a2;
	    for(a3=1;a3<=M;a3++){
		A[2]=a3;
		for(a4=1;a4<=M;a4++){
		    A[3]=a4;
		    for(a5=1;a5<=M;a5++){
			A[4]=a5;
			for(a6=1;a6<=M;a6++){
			    A[5]=a6;
			    for(a7=1;a7<=M;a7++){
				A[6]=a7;
				for(a8=1;a8<=M;a8++){
				    A[7]=a8;
				    for(a9=1;a9<=M;a9++){
					A[8]=a9;
					for(a10=1;a10<=M;a10++){
					    A[9]=a10;
					    newsum=0;
					    for(int i=0;i<Q;i++){
    						    if(A[abcd[4*i+1]-1]-A[abcd[4*i]-1]==abcd[4*i+2])
    							newsum+=abcd[4*i+3];
					    }
					    if(newsum>sum)
						sum=newsum;
					}
				    }	
				}			
			    }
			}
		    }
		}
	    }
	}
    }

    printf("%d\n",sum);

    return 0;
}