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 (Clang 10.0.0) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
 
int **dmatrix(int x,int y);	//x行y列の配列を確保する.
void free_dmatrix(int **a,int x);//x行の配列を解放する

int main(void){
	int N,M,Q;
	int **S,*A,max_p=0,p;
	int i;
	char str[100000];
 
	fgets(str, sizeof(str), stdin);
	N=atoi(strtok(str," "));
	M=atoi(strtok(NULL," "));
	Q=atoi(strtok(NULL," "));

	S=dmatrix(4,Q);
	A=(int*)calloc(N,sizeof(int));

	for(i=0;i<Q;i++){
		fgets(str, sizeof(str), stdin);	
		S[0][i]=atoi(strtok(str," "));	//a
		S[1][i]=atoi(strtok(NULL," "));	//b
		S[2][i]=atoi(strtok(NULL," ")); //c
		S[3][i]=atoi(strtok(NULL," "));	//d
	}

	for(i=0;i<N;i++){//初期化
		A[i]=1;
	}

	while(1){
		int f_num=1;

		/*得点計算*/
		p=0;
		for(i=0;i<Q;i++){
			if( (A[S[1][i]-1]-A[S[0][i]-1]) == S[2][i] ) p+=S[3][i];
		}
		if(p>max_p) max_p=p;
		/*得点計算*/
		A[N-1]=A[N-1]+1;
	
		if(A[N-1]>M){
			for(i=N-2;i>=0;i--){
				if( A[i] < M){
					A[i]=A[i]+1;
					f_num=A[i];
					break;
				}
			}
			if(i<0)break;
			i++;
			for(;i<N;i++){
				A[i]=f_num;
			}
		}
	}

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


	free_dmatrix(S,4);
	free(A);

	return 0;
}

int **dmatrix(int x,int y)			//x行y列の配列を確保する.
{
	int i;
	int **a;
	if ( ( a = (int**)calloc(x,sizeof(int*)) ) == NULL){
		//printf("メモリの確保に失敗しました.\nプログラムを終了します.");
		exit(1);
	}
	for(i=0;i<x;i++) a[i]=(int*)calloc(y,sizeof(int));
	return (a);
}

void free_dmatrix(int **a,int x)//x行の配列を解放する.
{
	int i;
	for (i=0;i<x;i++) free((void*)(a[i]));	
	free((void*)(a)); 
}