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 Java (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.Scanner;

/**
 * Main
 */
public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        int N = sc.nextInt();
        int M = sc.nextInt();
        int Q = sc.nextInt();
        int[] arrayA = new int[N];
        int[] a = new int[Q];
        int[] b = new int[Q];
        int[] c = new int[Q];
        int[] d = new int[Q];
        int sum = 0;
        int Max = 0;

        for(int i=0; i < N; i++){
            arrayA[i] = 1;
        }

        for(int i=0; i < Q; i++){
            a[i] = sc.nextInt();
            b[i] = sc.nextInt();
            c[i] = sc.nextInt();
            d[i] = sc.nextInt();
        }

        for(int i=N-1; i>=0; i--){
            while (arrayA[i]<M) {
                sum = 0;
                for(int j=0; j < Q; j++){
                   if(arrayA[b[j]-1]-arrayA[a[j]-1] == c[j]){
                       sum += d[j];
                   }
                }
                Max = Math.max(sum, Max);
                arrayA[i] += 1;
            }
        }
        System.out.println(Max);

    }
}