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>
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
using namespace std;
typedef long long ll;
int n,m,q;
vector<vector<int>> queries;
ll max_point = 0;

ll calc_point(vector<int> &lst){
    ll point = 0;
    rep(i,q){
        int a = queries.at(i).at(0)-1;
        int b = queries.at(i).at(1)-1;
        int c = queries.at(i).at(2);
        int d = queries.at(i).at(3);
        if(lst.at(b) - lst.at(a) == c){
            point += d;
        }
    }
    max_point = max(max_point, point);
    return point;
}

// N個の数列を全列挙
// a[i] <= a[i+1]
// a[i] = 1-m
void dfs(vector<int> lst, int i){
    if(i==n){
        // rep(j,lst.size()) cout << lst.at(j) << " ";
        // cout << endl;
        calc_point(lst);
        return;
    }

    int st;
    if(i==0){
        st = 1;
    }else {
        st = lst.at(i-1);
    }
    for(int j=st;j<=m;j++){
        lst.at(i) = j;
        dfs(lst, i+1);
    }
}

int main(){
    cin >> n >> m >> q;
    rep(i,q){
        vector<int> qrow(4);
        rep(j,4) cin >> qrow.at(j);
        queries.push_back(qrow);
    }
    vector<int> a(n);
    a[0] = 1;
    dfs(a,0);
    cout << max_point << endl;
    return 0;
}