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 <bits/stdc++.h>

using namespace std;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
#define all(x) x.begin(), x.end()

template <typename T1, typename T2> inline void chkmin(T1 &x, const T2 & y) {if (x > y) x = y;}
template <typename T1, typename T2> inline void chkmax(T1 &x, const T2 & y) {if (x < y) x = y;}

int n, m, q;
int a[50], b[50], c[50], d[50];
void read() {
    cin >> n >> m >> q;
    for (int i = 0; i < q; i++) {
        cin >> a[i] >> b[i] >> c[i] >> d[i];
    }
}

int ans = 0;

void relax(vector<int> & have) {
   /* cout << "have = " << endl;
    for (auto i : have)
        cout << i << " ";
    cout << endl; */
    int fans = 0;
    for (int i = 0; i < q; i++) {
        if (have[b[i] - 1] - have[a[i] - 1] == c[i]) {
            fans += d[i];
        }
    }
    chkmax(ans, fans);
}

void solve(vector<int> have) {
    if (have.size() == n) {
        relax(have);
        return;
    }
    if (have.empty()) {
        have.push_back(1);
    } else {
        have.push_back(have.back());
    }
    while (have.back() <= m) {
        solve(have);
        have[have.size() - 1]++;
    }
}

void run() {
    solve({});
}

void write() {
    cout << ans << endl;
}

signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    read();
    run();
    write();
    return 0;
}