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)

#pragma warning( disable : 6031)

#include <stdio.h>
#include <algorithm>
#include <vector>
#include <string>
#include <queue>
#include <map>
#include <math.h>
int minimum(int a, int b) { return a < b ? a : b; }
int maximum(int a, int b) { return a > b ? a : b; }
const long long mod = 1000000007;
//const long long mod = 998244353;
using std::vector;
using std::pair;
using std::string;
using std::map;
using std::abs;
int main() {
	int N, M, Q;
	scanf("%d %d %d", &N, &M, &Q);
	auto a = new int[Q];
	auto b = new int[Q];
	auto c = new int[Q];
	auto d = new int[Q];
	for (int i = 0; i < Q; i++)scanf("%d %d %d %d", a + i, b + i, c + i, d + i);
	int max = 0;
	auto A = new int[N] {};
	auto B = new int[N + M - 1]{};
	for (int i = 0; i < M - 1; i++)B[N + i] = 1;
	do {
		int j = 0;
		int k = 0;
		for (int i = 0; i < N + M - 1; i++) {
			if (B[i])j++;
			else {
				A[k] = j;
				k++;
			}
		}
		int sc = 0;
		for (int i = 0; i < Q; i++) {
			if (A[b[i] - 1] - A[a[i] - 1] == c[i]) {
				sc += d[i];
			}
		}
		//for (int i = 0; i < N; i++)printf(" %d ", A[i]);
		//printf("%d\n", sc);
		max = maximum(max, sc);
	} while (std::next_permutation(B, B + N + M - 1));
	printf("%d", max);
	return 0;
}