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)

#define _DEBUG 1
#define _GLIBCXX_DEBUG

#include <bits/stdc++.h>
using namespace std;

using ll = long long;

#ifdef _DEBUG
#define dump(x) cerr << #x << "=" << x << endl
#define dump2(x, y) cerr << #x << "=" << x << "," << #y << "=" << y << endl
#define dump3(x, y, z)                                                     \
  cerr << #x << "=" << x << "," << #y << "=" << y << "," << #z << "=" << z \
       << endl
#define check(s) cerr << s << endl
#else
#define dump(x)
#define dump2(x, y)
#define dump3(x, y, z)
#define check(s)
#endif

#define rep(i, n) for (int i = 0; i < n; i++)
#define repr(i, n) for (int i = n; i >= 0; i--)
#define FOR(i, m, n) for (int i = m; i < n; i++)
#define all(x) (x).begin(), (x).end()
#define sz(x) ((int)(x).size())
#define unique(v) v.erase(unique(v.begin(), v.end()), v.end());

const ll LINF = 2e18;
const int INF = 1e9;

ll N, M, Q;
vector<int> a, b, c, d;
int mx = -1;

void rec(vector<int> A, int idx) {
  // cerr << "A: ";
  // rep(i, N) { cerr << A.at(i) << " "; }
  // cerr << endl;

  int sum = 0;
  rep(i, Q) {
    if (A.at(b.at(i)) - A.at(a.at(i)) == c.at(i)) {
      sum += d.at(i);
    }
  }
  if (sum > mx) {
    mx = sum;
  }

  if (A.at(idx) < M) {
    vector<int> v = A;
    v.at(idx)++;
    rec(v, idx);

    if (idx != 0) {
      vector<int> w = A;
      w.at(idx - 1)++;
      rec(w, idx - 1);
    }
  } else {
    if (idx != 0) {
      rec(A, idx - 1);
    }
  }

  return;
}

void solve() {
  vector<int> A(N, 1);

  rec(A, N - 1);

  cout << mx << endl;
}

int main() {
  cin >> N >> M >> Q;
  a.resize(Q);
  b.resize(Q);
  c.resize(Q);
  d.resize(Q);
  rep(i, Q) {
    cin >> a.at(i) >> b.at(i) >> c.at(i) >> d.at(i);
    a.at(i)--;  // 0-indexed
    b.at(i)--;  // 0-indexed
  }
  solve();
  return 0;
}