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 <math.h>
#include <iostream>
#include <vector>
#include <algorithm>
#include <functional>
#include <queue>
#include <map>
#include <utility>
#include <iomanip>
using namespace std;

#define ll long long
ll N, M, Q;
vector<vector<ll>> LIST;

// n = 1, m = 0
void cal(ll n, ll m, vector<ll> XX)
{
    if (n == N)
    {
        LIST.push_back(XX);
        return;
    }

    for (ll i = m; i <= M; i++)
    {
        vector<ll> X2(XX.size());
        copy(XX.begin(), XX.end(), X2.begin());
        X2.push_back(i);
        cal(n + 1, i, X2);
    }
}

void solve()
{
    cin >> N >> M >> Q;
    vector<vector<ll>> ABCD;
    for (ll i = 0; i < Q; i++)
    {
        vector<ll> abcd(4);
        cin >> abcd.at(0) >> abcd.at(1) >> abcd.at(2) >> abcd.at(3);
        ABCD.push_back(abcd);
    }

    vector<ll> te;
    cal(0, 0, te);
    ll dcountmax = 0;
    for (auto A : LIST)
    {
        ll dcount = 0;
        // for (auto le : A)
        // {
        //     cout << le << " ";
        // }
        // cout << endl;

        for (ll j = 0; j < Q; j++)
        {
            if ((A.at(ABCD.at(j).at(1) - 1) - A.at(ABCD.at(j).at(0) - 1)) == ABCD.at(j).at(2))
            {
                dcount = dcount + ABCD.at(j).at(3);
            }
        }

        dcountmax = max(dcountmax, dcount);
    }

    cout << dcountmax << endl;

    cout << LIST.size() << endl;

    // cout << N << " " << M << " " << Q << endl;
    // for (ll j = 0; j < Q; j++)
    // {
    //     for (auto le : ABCD[j])
    //     {
    //         cout << le << " ";
    //     }
    //     cout << endl;
    // }

    // for (auto la : LIST)
    // {
    //     for (auto le : la)
    //     {
    //         cout << le << " ";
    //     }
    //     cout << endl;
    // }

    return;
}

int main()
{
    solve();
    return 0;
}