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;
using ll = long long;
typedef pair<int, int> P;
ll Mod = 1000000007;
int N,M,Q;
int a[55];
int b[55];
int c[55];
int d[55];
int ans = -1;
int res;

void DFS(vector<int> A) {
  int s = A.size();
  if (s == N) {
    res = 0;
    for (int i = 0; i < Q; i++) {
      if (A[b[i]] - A[a[i]] == c[i]) {
        res += d[i];
      }
    }
    ans = max(ans,res);
    return ;
  }
  for (int i = A[s-1]; i <= M; i++) {
    A.push_back(i);
    DFS(A);
    A.pop_back();
  }
  return ;
}

int main() {
  cin >> N >> M >> Q;
  for (int i = 0; i < Q; i++) {
    cin >> a[i] >> b[i] >> c[i] >> d[i];
    a[i]--;
    b[i]--;
  }
  for (int i = 1; i <= M; i++) {
    vector<int> X(0);
    X.push_back(i);
    DFS(X);
    X.pop_back();
  }
  cout << ans << endl;
  return 0;
}