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 <algorithm>
#include <cmath>
#include <vector>
#include <functional>
#include <cstdlib>
#include <map>
#include <set>
#include <iostream>
#include <string>
#include <ctype.h>
#include <climits>
#include <queue>
using namespace std;
#define REP(i, n) for(ll i = 0; i < (ll)(n); i++)
typedef long long ll;

int n,m,q;
vector<int> a,b,c,d;

int dfs(int depth, vector<int> &A){
	if(depth==n){
		int ans=0;
		REP(i,q){
			if(A[b[i]-1] - A[a[i]-1] == c[i]) ans += d[i];
		}
		return ans;
	}
	
	int ans=0;
	if(depth==0){
		for(int i=1; i<=m; i++){
			A[depth] = i;
			ans = max(ans, dfs(depth+1, A));
		}
		return ans;
	}else{
		for(int i=A[depth-1]; i<=m; i++){
			A[depth] = i;
			ans = max(ans, dfs(depth+1, A));
		}
		return ans;
	}
	
	
	
}

int main(){
	cin >> n >> m >> q;
	a.resize(q);
	b.resize(q);
	c.resize(q);
	d.resize(q);
	REP(i,q) cin >> a[i] >> b[i] >> c[i] >> d[i];
	
	vector<int> A(n);
	int ans = dfs(0,A);
	
	cout << ans << endl;
	
	return 0;
}