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)

// C.

#include <iostream>
#include <sstream>
#include <algorithm>
#include <numeric>
#include <vector>

using namespace std;

typedef pair<int, int> II;
typedef pair<II, II> IIII;
typedef long long LL;

int main(int argc, char* argv[]) {
	int n, m, q;
	cin >> n >> m >> q;
	vector<IIII> v(q);
	for (int i = 0; i < q; ++i) {
		cin >> v[i].first.first >> v[i].first.second >> v[i].second.first >> v[i].second.second;
		v[i].first.first -= 1;
		v[i].first.second -= 1;
	}
	vector<vector<int>> seq;
	for (int i = 1; i <= 10; ++i) {
		seq.emplace_back(vector<int>(1, i));
	}
	for (int i = 2; i <= n; ++i) {
		vector<vector<int>> next;
		for (auto v : seq) {
			for (int d = v.back(); d <= 10; ++d) {
				vector<int> w = v;
				w.emplace_back(d);
				next.emplace_back(w);
			}
		}
		seq = next;
	}
	LL ans = 0;
	for (auto a : seq) {
		LL score = 0;
		for (int i = 0; i < q; ++i) {
			if (a[v[i].first.second] - a[v[i].first.first] == v[i].second.first) {
				score += v[i].second.second;
			}
		}
		ans = max(ans, score);
	}
	cout << ans << endl;
	return 0;
}