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 <iomanip>
#include <string>
#include <map>
#include <set>
#include <vector>
#include <cmath>
#include <algorithm>
#include <numeric>
#include <bitset>
#include <unordered_map>
#include <unordered_set>
#include <random>
#include <queue>
#include <memory>
#include <fstream>

#pragma GCC optimize("no-stack-protector,Ofast,unroll-loops,fast-math")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,popcnt,abm,mmx,tune=native,avx,avx2")

#define int long long
#define ld long double
#define eb emplace_back
#define pb pop_back
#define mp make_pair
#define ss second
#define ff first
#define sqr(x) (x) * (x)
#define all(x) x.begin(), x.end()
#define sz(x) (int)x.size()
#define pii pair < int, int >
#define vi vector < int >
#define vvi(x, y) vector < vi > (x, vi(y))

typedef int(*func)(int a, int b);

using namespace std;

random_device rd;
mt19937 get_rand(rd());

template < typename T1, typename T2 >
istream& operator>>(istream& in, pair < T1, T2 >& p) {
	in >> p.ff >> p.ss;
	return in;
}

template < typename T1, typename T2 >
ostream& operator<<(ostream& out, pair < T1, T2 >& p) {
	out << p.ff << ' ' << p.ss;
	return out;
}

template < typename T1 >
istream& operator>>(istream& in, vector < T1 >& v) {
	for (int i = 0; i < sz(v); i++)
		in >> v[i];
	return in;
}

template < typename T1 >
ostream& operator<<(ostream& out, vector < T1 >& v) {
	for (int i = 0; i < sz(v); i++)
		out << v[i] << ' ';
	out << '\n';
	return out;
}

int gcd(int a, int b) {
	return (!a ? b : gcd(b % a, a));
}

const int INF = 1e9 + 7;
const int B = 100003;
const int MOD = 998244353;
const ld EPS = 1e-9;
const ld PI = 3.14159265358979323;
const int MAXLL = 9223372036854775807;

int a[50], b[50], c[50], d[50], A[50];
int n, m, q;
int res = 0;

void rec(int p, int k) {
	if (p == n) {
		int s = 0;
		for (int i = 0; i < q; i++)
			if (A[b[i] - 1] - A[a[i] - 1] == c[i])
				s += d[i];
		res = max(res, s);
		return;
	}
	for (int i = k; i <= m; i++) {
		A[p] = i;
		rec(p + 1, i);
	}
}

signed main() {
	//ifstream cin("kdivision.in");
	//ofstream cout("kdivision.out");
#ifdef _LOCAL
	freopen("output.txt", "w", stdout);
	freopen("input.txt", "r", stdin);
#endif
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cout << fixed << setprecision(10);
	srand(time(NULL));

	int ttt = 1;
	//cin >> ttt;
	while (ttt--) {
		cin >> n >> m >> q;
		for (int i = 0; i < q; i++) 
			cin >> a[i] >> b[i] >> c[i] >> d[i];
		rec(0, 1);
		cout << res << '\n';
	}
	return 0;
}