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;

// template {{{
#define  range(i, l, r) for (int i = (int)(l); i < (int)(r); (i) += 1)
#define rrange(i, l, r) for (int i = (int)(r) - 1; i >= (int)(l); (i) -= 1)

#define  whole(f, x, ...) ([&](decltype((x)) container) { return (f)(  begin(container),  end(container), ## __VA_ARGS__); })(x)
#define rwhole(f, x, ...) ([&](decltype((x)) container) { return (f)( rbegin(container), rend(container), ## __VA_ARGS__); })(x)

#define debug(x) cerr << "(" << __LINE__ << ")" << #x << ": " << (x) << endl

using i32 = int;
using u32 = unsigned int;
using i64 = long long;
using u64 = unsigned long long;

// constexpr i32 mod   = 998244353;
constexpr i32 mod   = 1e9 + 7;
constexpr i32 inf   = 1001001001;
constexpr i64 infll = 1001001001001001001ll;

constexpr int dx[] = {0, -1, 1, 0, -1, 1, -1, 1};
constexpr int dy[] = {-1, 0, 0, 1, -1, -1, 1, 1};

struct IoSetup { IoSetup(int x = 15){ cin.tie(0); ios::sync_with_stdio(0); cout << fixed << setprecision(x); cerr << fixed << setprecision(x); } } iosetup;

template <typename T = i64> T input() { T x; cin >> x; return x; }

template <typename T> ostream &operator<<(ostream &os, vector<T> &v) { range(i, 0, v.size()) { os << v[i] << (i + 1 != v.size() ? " " : ""); } return os; }
template <typename T> istream &operator>>(istream &is, vector<T> &v) { for (T &in : v) is >> in; return is; }

template <typename T1, typename T2> ostream &operator<<(ostream &os, pair<T1, T2> p) { os << "(" << p.first << ", " << p.second << ")"; return os; }
template <typename T1, typename T2> istream &operator>>(istream &is, pair<T1, T2> &p) { is >> p.first >> p.second; return is; }

template <typename T> vector<T> make_vector(size_t a, T b) { return vector<T>(a, b); }
template <typename... Ts> auto make_vector(size_t a, Ts... ts) { return vector<decltype(make_vector(ts...))>(a, make_vector(ts...)); }

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); }
// }}}

void solver() {
    int n = input(), m = input(), q = input();
    vector< int > as(q), bs(q), cs(q), ds(q);
    range(i, 0, q) {
        cin >> as[i] >> bs[i] >> cs[i] >> ds[i];
    }

    auto calc = [&](vector< int > A) {
        i64 res = 0;
        range(i, 0, q) {
            if (A[bs[i] - 1] - A[as[i] - 1] == cs[i]) res += ds[i];
        }
        return res;
    };

    int ans = 0;
    vector< int > A(n);
    function<void(int, int)> dfs = [&](int idx, int pre) {
        if (idx == n) {
            chmax(ans, calc(A));
            return;
        }
        range(i, pre, m + 1) {
            A[idx] = i;
            dfs(idx + 1, i);
        }
    };

    dfs(0, 0);
    cout << ans << endl;
}

signed main(int argc, char *argv[]) {
    solver();
}