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>
using namespace std;
using ll = long long;

const int MOD = 1e9+7;
int add(int a, int b){return ((ll)a + (ll)b)%MOD;}
int mul(int a, int b){return ((ll)a * (ll)b)%MOD;}

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

    ll ans = 0;
    for(int i = 0; i < 1<<(n); i++){
        ll poss = 0;
        vector<int> A;
        for(int j = 0; j <= n; j++){
            if(i>>j & 1){
                A.push_back(j);
            }
        }
        //if(a.size() != min(n, m)){continue;}
        for(int j = 0; j < q; j++){
            if(b[j] < A.size() && a[j] < A.size() && A[b[j]] - A[a[j]] == c[j]){
                poss += d[j];
            }
        }
        // if(poss){
        //     cout << poss << ": "; 
        //     for(int i : A){
        //     cout << i+1 << ' ';
        // }cout << endl;
        // }  
        ans = max(ans, poss);
    }
    cout << ans << endl;
}

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    solve();
    return 0;
}