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 11.0.6) (judge verdict: Wrong Answer)

import java.util.Scanner;

public class Main{
	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		
		int n = scan.nextInt();
		int m = scan.nextInt();
		int q = scan.nextInt();
		int[] a = new int[q];
		int[] b = new int[q];
		int[] c = new int[q];
		int[] d = new int[q];
		for(int i = 0; i < q; i++) {
			a[i] = scan.nextInt();
			b[i] = scan.nextInt();
			c[i] = scan.nextInt();
			d[i] = scan.nextInt();
		}
		scan.close();
		
		System.out.println(dfs(a,b,c,d,new int[n],m,0,0));
	}
	
	public static int dfs(int[] a, int[] b, int[] c, int[] d, int[] num, int m, int i, int prev) {
		if(i == num.length) {
			return Ans(a,b,c,d,num);
		}
		
		int[] candidate = new int[m-prev+1];
		for(int j = 0; j < m-prev+1; j++) {
			num[i] = prev+j;
			candidate[j] = dfs(a,b,c,d,num,m,i+1,prev+j);
		}
		
		return max(candidate);
	}
	

	
	public static int Ans(int[] a, int[] b, int[] c, int[] d, int[] num) {
		int ans = 0;
		for(int i = 0; i < a.length; i++) {
			if(num[b[i]-1] - num[a[i]-1] == c[i]) {
				ans += d[i];
			}
		}
		return ans;
	}
	
	public static int max(int[] candidate) {
		int max = candidate[0];
		for(int i = 1; i < candidate.length; i++) {
			if(candidate[i] > max) max = candidate[i];
		}
		return max;
	}
}