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>
#define ios ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0)
#define debug  freopen("in.txt","r",stdin),freopen("out.txt","w",stdout);
#define pb push_back
#define all(x) x.begin(),x.end()
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int maxn = 1e6+10;
const int maxM = 1e6+10;
const int inf = 0x3f3f3f3f;

int N,M,Q;
struct node
{
	int a,b,c,d;
}qq[maxn];
int A[maxn];
int res = 0;
void dfs(int id,int last = 1){
	if(id > N){
		int sum = 0;
		for(int i = 1;i<=Q;i++){
			if(A[qq[i].b] - A[qq[i].a] == qq[i].c)
				sum += qq[i].d;
		}
		res = max(res,sum);
	}else{
		for(int i = last;i<=M;i++){
			A[id] = i;
			dfs(id+1,i);
		}
	}
}
int main(){
	// debug;
	ios;

	cin>>N>>M>>Q;
	for(int i = 1;i<=Q;i++){
		cin>>qq[i].a >> qq[i].b >> qq[i].c>>qq[i].d;
	}
	dfs(1);
	cout<<res<<'\n';

	return 0;
}