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 <stdio.h>
#include <stdlib.h>
#include <sstream>
#include <iostream>
#include <vector>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
using namespace std;
#define fio ios_base::sync_with_stdio(false); cin.tie(NULL);
#define M 1000000007
//#define FILE_A_IN
using vvi=vector<vector<int>>;

int n, m, q;
void dfs(vector<vector<int>>& va, vector<int> vt, int next, int& ans) {
    if(vt.size() == n) {
        int s = 0;
        for (auto e : va) {
            if(vt[e[1]-1] - vt[e[0]-1] == e[2]) {
                s += e[3];
            }
        }
        ans = max(s, ans);
        return;
    }
    for (int i = next; i < m+1; ++i) {
        vt.push_back(i);
        dfs(va, vt, i, ans);
        vt.pop_back();
    }
}
void solve() {
    int ans = 0;
    cin >> n >> m >> q;
    vvi va;
    for (int i = 0; i < q; ++i) {
        vector<int> vt(4);
        for (int j = 0; j < 4; j++) {
            cin >> vt[j];
        }
        va.push_back(vt);
    }
    vector<int> vt;
    dfs(va, vt, 1, ans);
    cout << ans << endl;
}
int main() {
#ifdef FILE_A_IN
    freopen("a.in", "r", stdin);
#endif
    fio;
        solve();
    return 0;
}