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 <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0;i<(n);i++)
#define rep1(i,n) for(int i=1;i<=(n);i++)
#define all(x) (x).begin(),(x).end()
#define V vector
typedef V<int> vi;
typedef V<vi> vvi;
typedef long long ll;
typedef pair<ll, ll> P;
constexpr auto INF = INT_MAX / 2;
constexpr auto LINF = 5000000000000000;
constexpr auto MOD = 1000000007;

int main() {

	int n, m, q;
	cin >> n >> m >> q;
	vi a(q), b(q), c(q), d(q);
	rep(i, q) {
		cin >> a[i] >> b[i] >> c[i] >> d[i];
		a[i];b[i];
	}

	int ans = 0;

	for (ll i = 0;i < (1LL << (n + m + 1));i++) {
		int hoge = 0;
		rep(j, n + m + 1)if (i >> j & 1)hoge++;
		if (hoge != m - 1)continue;
	

		vi A(n + 2);
		hoge = 0;
		int ai = 1;

		rep(j, n + m + 1) {
			if (i >> j & 1)ai++;
			else { A[hoge] = ai;hoge++; }
		}

		int score = 0;

		rep(j, q)if (A[b[j]] - A[a[j]] == c[j])score += d[j];
		
		ans = max(ans, score);

	}

	cout << ans << endl;

}