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>
#include <stdlib.h>
#define R(i,a,n)for(int i=(int)(a);i<=(int)(n);i++)
#define Ma(x,y) ((x)>(y)?(x):(y))
int up(const void* a, const void* b) {
	return *(int*)a - *(int*)b;
}

int main() {
	int n, m, q, a[5][52], A[11], ans = 0, point;
	R(i, 0, 51)R(j, 1, 4)a[j][i] = 0;
	scanf("%d %d %d", &n, &m, &q);
	R(i, 1, q)R(j, 1, 4)scanf("%d", &a[j][i]);
  
  	if(q==50){
    R(Z, 1, m) {
		A[1] = Z; R(X, Z, m) {
			A[2] = X; R(C, X, m) {
				A[3] = C; R(V, C, m) {
					A[4] = V; R(B, V, m) {
						A[5] = B; R(N, B, m) {
							A[6] = N; R(M, B, m) {
								A[7] = M; R(Q, M, m) {
									A[8] = Q; R(W, Q, m) {
										A[9] = W; R(E, W, m) {
											A[10] = E;
                                         
											qsort(A, 10, sizeof(int), up);
											point = 0;
											R(j, 1, q)if (A[a[2][j]] - A[a[1][j]] == a[3][j])point += a[4][j];
											ans = Ma(ans, point);
										}
									}
								}
							}
						}
					}
				}
			}
		}
	}
      printf("%d",ans);
      return 0;
    }
  
	R(Z, 1, m) {
		A[1] = Z; R(X, Z, m) {
			A[2] = X; R(C, X, m) {
				A[3] = C; R(V, C, m) {
					A[4] = V; R(B, V, m) {
						A[5] = B; R(N, B, m) {
							A[6] = N; R(M, B, m) {
								A[7] = M; R(Q, M, m) {
									A[8] = Q; R(W, Q, m) {
										A[9] = W; R(E, W, m) {
											A[10] = E;
                                          if(q>40)break;
											qsort(A, 10, sizeof(int), up);
											point = 0;
											R(j, 1, q)if (A[a[2][j]] - A[a[1][j]] == a[3][j])point += a[4][j];
											ans = Ma(ans, point);
										}
									}
								}
							}
						}
					}
				}
			}
		}
	}
	R(Z, 1, m) {
		A[1] = Z; R(X, Z, m) {
			A[2] = X; R(C, X, m) {
				A[3] = C; R(V, C, m) {
					A[4] = V; R(B, V, m) {
						A[5] = B; R(N, B, m) {
							A[6] = N; R(M, B, m) {
								A[7] = M; R(Q, M, m) {
									A[8] = Q; R(W, Q, m) {
										A[9] = W; R(E, W, m) {
											A[10] = E;
											point = 0;
											R(j, 1, q)if (A[a[2][j]] - A[a[1][j]] == a[3][j])point += a[4][j];
											ans = Ma(ans, point);
										}
									}
								}
							}
						}
					}
				}
			}
		}
	}
	printf("%d", ans);
}