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>
#define REP(i, n) for (int i = 0; i < n; ++i)
#define REPR(i, n) for (int i = n - 1; i >= 0; --i)
#define FOR(i, a, b) for (int i = a; i < b; ++i)
#define FORR(i, a, b) for (int i = b - 1; i >= a; --i)
#define SORT(v) sort(v.begin(), v.end())
#define SORTR(v) sort(v.rbegin(), v.rend())
#define REV(v) reverse(v.begin(), v.end())
#define ITER(itr, v) for (auto itr = v.begin(); itr != v.end(); ++itr)
#define LB(v, x) (lower_bound(v.begin(), v.end(), x) - v.begin())
#define UB(v, x) (upper_bound(v.begin(), v.end(), x) - v.begin())
#define SZ(v) (int)v.size()
using namespace std;
using ll = long long;
using P = pair<int, int>;

int N, M, Q;
int a[51], b[51], c[51], d[51];
void dfs(int i, int x, int &res, vector<int> &A) {
    if (i == N) {
        int sum = 0;
        REP(j, Q) {
            if (A[b[j]] - A[a[j]] == c[j]) sum += d[j];
        }
        res = max(res, sum);
        return;
    }
    A[i] = x;
    for (int y = x; y <= M; ++y) { dfs(i + 1, y, res, A); }
}

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

    cin >> N >> M >> Q;
    REP(i, Q) {
        cin >> a[i] >> b[i] >> c[i] >> d[i];
        --a[i];
        --b[i];
    }
    int res = 0;
    vector<int> A(N);

    FOR(x, 1, M) { dfs(0, x, res, A); }
    cout << res << endl;

    return 0;
}