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 <iostream>
#include <vector>
using namespace std;
struct Query {
    int a, b, c, d;
    Query(int aa = 0, int bb = 0, int cc = 0, int dd = 0)
    : a(aa), b(bb), c(cc), d(dd) {}
};

Query qs[55];
int N, M, Q;
long long ans = 0;
int nums[12];
vector<int> vec;
void dfs(int i) {
    if (i == M + 1) {
        if (vec.size() == N + 1) {
            long long summ = 0;
            for (int k = 1; k <= Q; ++k) {
                if (vec[qs[k].b] - vec[qs[k].a] == qs[k].c) {
                    summ += qs[k].d;
                }
            }
            ans = max(ans, summ);
        } 
        return ;
    }
    for (int j = 1; j <= M; ++j) {
        if (j >= vec[i - 1]) {
            vec.push_back(j);
            dfs(i + 1);
            vec.pop_back();
            dfs(i + 1);
        }
    }
}

int main() {
    ios::sync_with_stdio(0);
    cin >> N >> M >> Q;
    for (int i = 1; i <= Q; ++i) {
        cin >> qs[i].a >> qs[i].b >> qs[i].c >> qs[i].d;
    }
    for (int i = 1; i <= M; ++i)
        nums[i] = i;
    vec.push_back(0);
    dfs(1);
    cout << ans << endl;
    return 0;
}