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: Accepted)

import java.util.*;
import static java.lang.Math.*;

public class Main {
	public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner(System.in);
		
		// 入力
		int n = sc.nextInt();
		int m = sc.nextInt();
		int q = sc.nextInt();
		int[][] query = new int[q][4];
		for(int i = 0; i < q; i++){
		    query[i][0] = sc.nextInt()-1;
		    query[i][1] = sc.nextInt()-1;
		    query[i][2] = sc.nextInt();
		    query[i][3] = sc.nextInt();
		}
		
		// 計算
		long result = 0;
		int[] list = new int[n];
		result = func(list, m, 0, 0, query);
		
		// 出力
		System.out.println(result);
	}
	
	static int func(int[] list, int m, int i, int result, int[][] query){
	    if(list.length == i){
	        return calc(list, query);
	    }
	    int from = 1;
	    if(i > 0) from = list[i-1];
	    for(int j = from; j <= m; j++){
	        list[i] = j;
	        result = max(result, func(list, m, i+1, result, query));
	    }
	    return result;
	}
	
	static int calc(int[] list, int[][] query){
	    int sum = 0;
	    for(int i = 0; i < query.length; i++){
	        if(list[query[i][1]] - list[query[i][0]] == query[i][2]){
	            sum += query[i][3];
	        }
	    }
	    return sum;
	}
}