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>
using namespace std;

#define MOD (1000000000+7)
#define pb(x) push_back(x)
#define mp(x,y) make_pair(x,y)
#define all(x) x.begin(), x.end()
#define print(vec,l,r) for(int i = l; i <= r; i++) cout << vec[i] <<" "; cout << endl;
#define forf(i,a,b) for(int i = (a); i < (b); i++)
#define forr(i,a,b) for(int i = (a); i > (b); i--)
#define input(vec,N) for(int i = 0; i < (N); i++) cin >> vec[i];

// template starts
typedef long long int ll;
int ans = 0;
vector<int> nums;
int N, M, Q;
vector<vector<int>> vec;
int score(){
	int s = 0;
	for(int i = 0; i < Q; i++){
		int b = vec[i][1]-1, a = vec[i][0]-1, c = vec[i][2], d = vec[i][3];
		if(nums[b] -nums[a] == c) s+= d;
	}
	return s;
}

void f(int i, int prev){
	if(i == N){
//		cout << endl;
//		for(int j = 0; j <N; j++) cout<< nums[j] << " ";
//		cout << " = " << score();
//		cout << endl;
		
		ans = max(ans, score());
		return;
	}
	for(int j = prev; j <= M; j++){
		nums[i] = j;
		f(i+1,j);
	}
}
void solve(){
	// code starts from here
	
	cin >> N >> M >> Q;
	vec.resize(Q);
	for(int q = 0; q < Q; q++){
		vec[q].resize(4);
		for(int i = 0; i < 4; i++){
			cin >> vec[q][i];
		}
	}
	
	nums.resize(N);
	for(int i = 0; i < N; i++) nums[i] = 0;
	
	f(0,1);
	cout << ans << endl;
}

int main(){
 	ios_base::sync_with_stdio(false);
    cin.tie(NULL);

	int T;
	T = 1;
	while(T--){
		solve();
	}

	return 0;
}