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 <set>
#include <map>
#include <algorithm>
#include <cmath>
#include <numeric>
#include <string>
#include <queue>
#include <iomanip>
#include <iterator>

#define MOD 1000000007
#define INF 1000000001

using namespace std;
typedef unsigned long long ull;
typedef long long ll;


template <class T>
ostream &operator<<(ostream &o, const vector<T>&obj) {
    o << "["; for (int i = 0; i < (int)obj.size(); ++i) o << (i > 0 ? ", " : "") << obj[i]; o << "]"; return o;
}
template <class T, class U>
ostream &operator<<(ostream &o, const pair<T, U>&obj) {
    o << "(" << obj.first << ", " << obj.second << ")"; return o;
}
template <class T, class U>
ostream &operator<<(ostream &o, const map<T, U>&obj) {
    o << "{"; for (auto itr = obj.begin(); itr != obj.end(); ++itr) o << (itr != obj.begin() ? ", " : "") << *itr; o << "}"; return o;
}
template <class T>
ostream &operator<<(ostream &o, const set<T>&obj) {
    o << "{"; for (auto itr = obj.begin(); itr != obj.end(); ++itr) o << (itr != obj.begin() ? ", " : "") << *itr; o << "}"; return o;
}
template <class T>
ostream &operator<<(ostream &o, const queue<T>&obj) {
    queue<T> obj_cp = obj;
    bool first = true;
    o << "{";
    while (!obj_cp.empty()) { o << (!first ? ", " : "") << obj_cp.front(); obj_cp.pop(); first = false; }
    o << "}"; return o;
}

void print() {
    cout << "\n";
}
template<class Head, class... Body>
void print(Head head, Body... body) {
    cout << head << " ";
    print(body...);
}


int N, M, Q;
vector<int> a, b, c, d;

int calc(const vector<int>& A) {
    int ret = 0;
    for (int i = 0; i < Q; ++i) {
        if (A[b[i]] - A[a[i]] == c[i]) {
            ret += d[i];
        }
    }
    return ret;
}

int search(int depth, vector<int> A) {
    if (depth == N) {
        // print(A, calc(A));
        return calc(A);
    }

    int ret = 0;
    int start = 1;
    if (depth > 0) {
        start = A[depth-1];
    }
    for (int i = start; i <= M; ++i) {
        A[depth] = i;
        ret = max(search(depth + 1, A), ret);
    }
    return ret;
}

int main(int argc, char const *argv[]) {
    cin >> N >> M >> Q;
    a.resize(Q);
    b.resize(Q);
    c.resize(Q);
    d.resize(Q);
    for (int i = 0; i < Q; ++i) {
        cin >> a[i] >> b[i] >> c[i] >> d[i];
        a[i] -= 1;
        b[i] -= 1;
    }

    vector<int> A(N, 0);
    int ans = search(0, A);
    cout << ans << endl;
    
    return 0;
}