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<iostream>
#include<string>
#include <algorithm>
#include<vector>
using namespace std;

/* Many Requirements */

int main(){
    // 入力スタンダード
    int n, m, q;
    cin >> n >> m >> q;

    vector<int> a;
    vector<int> b;
    vector<int> c;
    vector<int> d;

    a.push_back(0);
    b.push_back(0);
    c.push_back(0);
    d.push_back(0);

    for(int i=0; i<q; i++) {
        int ai,bi,ci,di;
        cin >> ai >> bi >> ci >> di;
        a.push_back(ai);
        b.push_back(bi);
        c.push_back(ci);
        d.push_back(di);
    }
    // ***************************************

    int A[n+1];
    for (int i=0; i<n+1; i++){
        A[i] =1;
    }
    
    int sumd = 0;
    int maxsum = 0;
    bool flag = true;

    while(flag){

        for(int i=1; i<q+1; i++){
            if(A[b[i]]-A[a[i]] == c[i]){
                sumd += d[i];
            }
        }

        if(maxsum < sumd){
            maxsum = sumd;
        }
       
        A[n] += 1;
        for(int j=n; j>0; j--){
            if(A[j]>m){
                A[j-1] += 1;
                for(int k=j; k<n+1; k++){
                    A[k] = A[j-1];
                }
                if(A[0]==2){
                    flag = false;
                }
            }
        }
        sumd = 0;
    }    

    cout << maxsum << endl;

    return 0;
}