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 ll long long
#define pll pair<long long,long long>
#define ii pair<int,int>
#define mod 1000000007
#define endl '\n'
#define IOS std::ios::sync_with_stdio(false); cin.tie(NULL);cout.tie(NULL);

ll power(ll a,ll b){
	ll ans=1;
	while(b>0){
		if(b%2!=0){
			ans=(ans*a)%mod;
		}
		a=((a%mod)*(a%mod))%mod;
		b>>=1;
	}
	return ans;
}

void sieve(int n){
	bool prime[n+1];
	memset(prime,true,sizeof(prime));
	for(int p=2;p*p<=n;p++){
		if(prime[p]){
			for(int i=p*p;i<=n;i+=p)
				prime[i]=false;
		}
	}
	for(int p=2;p<=n;p++){
		if(prime[p])
			cout<<p<<" ";
	}
}

void ans(int i,int last,int n,int m,int q,int &ma,int a[],pair<pair<int,int>,pair<int,int>> p[]){
	if(i==n){
		int ha=0;
		for(int j=0;j<q;j++){
			if(a[p[j].first.second-1]-a[p[j].first.first-1]==p[j].second.first)
			ha+=p[j].second.second;
		}
		ma=max(ha,ma);
		return;
	}
	for(int j=last;j<=m;j++){
		a[i]=j;
		ans(i+1,j,n,m,q,ma,a,p);
	}
}

int main() {
	IOS;
	int t=1;
	//cin>>t;
	while(t--){
		ll n,m,q;
		cin>>n>>m>>q;
		pair<pair<int,int>,pair<int,int>> p[q];
		for(int i=0;i<q;i++)
		cin>>p[i].first.first>>p[i].first.second>>p[i].second.first>>p[i].second.second;
		int a[n];
		memset(a,0,sizeof(a));
		int ma=0;
		ans(0,1,n,m,q,ma,a,p);
		cout<<ma;
	}
	return 0;
}