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)

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

// repetition
#define FOR(i, a, b) for(ll i = (a); i < (b); ++i)
#define rep(i, n) for(ll i = 0; i < (ll)(n); i++)

// container util
#define all(x) (x).begin(), (x).end()

// typedef
typedef long long ll;
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<ll> VLL;
typedef vector<VLL> VVLL;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef pair<ll, ll> PLL;

// const value
// const ll MOD = 1e9 + 7;
// const int dx[] = {0,1,0,-1};//{0,0,1,1,1,-1,-1,-1};
// const int dy[] = {1,0,-1,0};//{1,-1,0,1,-1,0,1,-1};

// conversion
inline int toInt(string s) {
  int v;
  istringstream sin(s);
  sin >> v;
  return v;
}
inline ll toLL(string s) {
  ll v;
  istringstream sin(s);
  sin >> v;
  return v;
}
template <class T> inline string toString(T x) {
  ostringstream sout;
  sout << x;
  return sout.str();
}

template <typename T1, typename T2> inline bool chmax(T1 &a, T2 b) {
  return a < b && (a = b, true);
}

template <typename T1, typename T2> inline bool chmin(T1 &a, T2 b) {
  return a > b && (a = b, true);
}

int n, m, q;
ll a[10], b[10], c[10], d[10];
ll num[10];
ll dfs(int idx) {
  ll res = 0;
  if(idx == n) {
    rep(j, q) {
      if(num[b[j]] - num[a[j]] == c[j]) {
        res += d[j];
      }
    }
    return res;
  }
  int start = 1;
  if(idx != 0)
    start = num[idx - 1];
  FOR(i, start, m + 1) {
    num[idx] = i;
    chmax(res, dfs(idx + 1));
  }
  return res;
}

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

  cin >> n >> m >> q;

  rep(i, q) {
    cin >> a[i] >> b[i] >> c[i] >> d[i];
    a[i]--, b[i]--;
  }

  cout << dfs(0) << endl;
  return 0;
}