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: Accepted)

#include <iostream>
#include <vector>
#include <map>
#include <algorithm>
#include <math.h>
#include <set>

using namespace std;

void create_comb(vector<vector<long long>> &arr, int n, int m, vector<long long> arr_temp) {

	int i;

	if (int(arr_temp.size()) == n) {
		arr.push_back(arr_temp);
		return;
	}

	for (i = arr_temp[arr_temp.size() - 1]; i <= m; i++) {
		arr_temp.push_back(i);

		create_comb(arr, n, m, arr_temp);

		arr_temp.pop_back();
	}

	return;
}


int main(void) {
	int n, m, q;
	long long i, j, k;
	long long temp_a, temp_b, temp_c, temp_d;
	vector<long long> a, b, c, d;
	vector<vector<long long>> arr;
	vector<long long> arr_temp;
	long long max = 0;
	long long sum = 0;

	cin >> n >> m >> q;

	for (i = 0; i < q; i++) {
		cin >> temp_a >> temp_b >> temp_c >> temp_d;
		a.push_back(temp_a);
		b.push_back(temp_b);
		c.push_back(temp_c);
		d.push_back(temp_d);
	}

	arr_temp.push_back(1);
	create_comb(arr, n, m, arr_temp);

	for (i = 0; i < arr.size(); i++) {
		for (j = 0; j < q; j++) {
			if (arr[i][b[j] - 1] - arr[i][a[j] - 1] == c[j]) {
				sum += d[j];
			}
		}

		if (max < sum) {
			max = sum;
		}

		sum = 0;
	}

	cout << max << endl;

	return 0;
}