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: Wrong Answer)

#include<bits/stdc++.h>
#include<algorithm>
using namespace std;
#define cool ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL) 
#define ll long long
int main()
{ 
    cool;
    ll int n,m,q,maxi=0;
    cin>>n>>m>>q;
    vector< pair<ll int,vector<ll int>> > v;
    for(ll int i=0;i<q;i++){
    	ll int a,b,c,d;
    	cin>>a>>b>>c>>d;
    	vector<ll int> v1;
    	v1.push_back(a);
    	v1.push_back(b);
    	v1.push_back(c);
    	v.push_back(make_pair(d,v1));
	}
	sort(v.begin(),v.end());
	for(ll int i=q-1;i>=0;i--){
		ll int sum=0;
		ll int a[n+1];
		for(ll int j=1;j<=n;j++){
			a[j]=0;
		}
		for(int j=i;j>=0;j--){
			vector<ll int> v1;
			v1=v[j].second;
			if(a[v1[0]]==0||a[v1[1]]==0){
			if(a[v1[0]]==0)
				a[v1[0]]=v1[0];
			if(a[v1[1]]==0)
				a[v1[1]]=a[v1[0]]+v1[2];	
			}
			if((a[v1[1]]-a[v1[0]])==v1[2]){
				sum+=v[j].first;
			}
		}
		maxi=max(maxi,sum);
				
	}
	cout<<maxi;
    return 0;
}