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 <vector>
#include <queue>
#include <algorithm>
#include <cmath>
#include <bitset>
#include <iomanip>
#include <stack>
#include <list>
#include <map>
#include <unordered_map>
#include <chrono>
#include <numeric>
using namespace std;
using ll = long long;
const ll INF = (ll)1e18+1;
const ll DIV = 1000000007;
//#define TEST

static bool COUNTUP = false;
bool countup(vector<ll>& A, const ll &end, ll depth)
{
    bool loop_end = true;
    //最初はカウントアップしない
    if(!COUNTUP)
    {
        COUNTUP=true;
        return true;
    }
    if(A[depth]< end) A[depth]++;
    else
    {
        if (depth == 0) return false;
        loop_end = countup(A, end, depth-1);
        A[depth] = A[depth-1];
    }
    return loop_end;
}

int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
#ifdef TEST
    chrono::system_clock::time_point start, end;
    start = chrono::system_clock::now();
#endif
    ll N, M, Q;
    cin >> N >> M >> Q;
    ll a[Q], b[Q], c[Q], d[Q];
    ll max_total = 0;
    vector<ll> A(N,1);
    for(size_t i=0;i<Q; i++)
    {
        cin >> a[i] >> b[i] >> c[i] >> d[i];
    }
    while(countup(A, M, A.size()-1))
    {
        ll total = 0;
        for(size_t i=0;i<Q; i++)
        {
            if(A[b[i]-1] - A[a[i]-1] == c[i]) total += d[i];
        }
        max_total = max(max_total, total);
    }
    cout << max_total << endl;
#ifdef TEST
    end = chrono::system_clock::now();
    cerr << static_cast<double>(chrono::duration_cast<chrono::microseconds>(end - start).count() / 1000.0) << "[ms]" << endl;
#endif
    return 0;
}