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>
#include<algorithm>
#include<iostream>
using namespace std;
using ll = long long;
#define rep(i,n) for(ll i=0;i<(ll)n;++i)
using P = pair<ll,ll>;

ll n,m,q;
vector<ll> a,b,c,d;
ll ans = -1;

void dfs( int s, int now, vector<int> v ){
	if( now == n+1 ){
		ll cnt = 0;
		rep(i,q){
			if( v.at(b.at(i)) - v.at(a.at(i)) == c.at(i) ){
				cnt += d.at(i);
			}
		}
		ans = max(ans,cnt);
		return;
	}
	for(int i=s;i<=m;++i){
		vector<int> tmp = v;
		tmp.emplace_back(i);
		dfs(i,now+1,tmp);
	}
}

int main(){
	cin >> n >> m >> q;
	a.resize(q,0);
	b.resize(q,0);
	c.resize(q,0);
	d.resize(q,0);
	rep(i,q){
		cin >> a.at(i) >> b.at(i) >> c.at(i) >> d.at(i);
		a.at(i)--;
		b.at(i)--;
	}
	vector<int> v = {};
	dfs(1,1,v);
	cout << ans << endl;
	return 0;
}