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 <algorithm>
#include <utility>
#include <string>
#include <queue>
#include <stack>

using namespace std;

typedef long long int ll;
typedef pair<int, int> Pii;

const ll mod = 1000000007;

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);

  int n, m, q;
  cin >> n >> m >> q;
  vector<vector<int> > abcd(q, vector<int>(4));
  for (auto &x: abcd) {
    cin >> x[0] >> x[1] >> x[2] >> x[3];
    x[0]--;
    x[1]--;
  }

  queue<vector<int> > que;
  for (int i = 0; i < m; i++) que.push(vector<int>({i}));
  ll ans = 0;
  while (!que.empty()) {
    auto now = que.front();
    que.pop();
    if ((int) now.size() < n) {
      for (int i = now[now.size()-1]; i < m; i++) {
        auto next = now;
        next.push_back(i);
        que.push(next);
      }
    }
    else {
      ll score = 0;
      for (int i = 0; i < q; i++) {
        if (now[abcd[i][1]] - now[abcd[i][0]] == abcd[i][2]) score += abcd[i][3];
      }
      if (score > ans) ans = score;
    }
  }

  cout << ans << endl;

  return 0;
}