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 endl '\n'
#define pb push_back
#define mp make_pair
#define SZ(x) ((int)(x).size())
#define all(x) x.begin(),x.end()
#define mem0(s) memset(s,0,sizeof(s))
#define meminf(s) memset(s,0x3f,sizeof(s))
#define mp_insert(a,b,c) mp.insert(map<pair<int,int>,int>::value_type(make_pair(a,b),c))
double pi=acos(-1);
typedef long long ll;
typedef unsigned long long ull;
inline ll lowbit(ll x){return x&(-x);}
inline ll min(ll a,ll b){return a<b?a:b;}
inline ll max(ll a,ll b){return a>b?a:b;}
const ll INF=4223372036854775807,max_n=300005,mod=0x7f7f7f7f7f;
inline ll gcd(ll a,ll b){return b==0?a:gcd(b,a%b);}
inline ll pow(ll a,ll b,ll mod){ll res=1;a%=mod;assert(b>=0);for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
struct Node
{
	int a,b,c,d;
};
Node data1[105];
bool cmp(Node a,Node b)
{
	return a.d>b.d;
}
int n,m,q;
int a[15]={0,1,1,1,1,1,1,1,1,1,1};
ll max1;
ll sum1;
void check()
{
	for(int i=1;i<=q;i++)
	{
		if(a[data1[i].b]-a[data1[i].a]==data1[i].c)
		{
			sum1+=data1[i].d;
		}
	}
	max1=max(max1,sum1);
	sum1=0;
	return;
}
void dfs(int x)
{
	if(x==n+1)
	{
		check();
		return;
	}
	for(int i=a[x-1];i<=m;i++)
	{
		a[x]=i;dfs(x+1);
	}
	
}
int main()
{
	cin>>n>>m>>q;
	for(int i=1;i<=q;i++)
	{
		cin>>data1[i].a>>data1[i].b>>data1[i].c>>data1[i].d;
	}
	//sort(data1+1,data1+1+q,cmp);
	//(int i=1;i<=q;i++)
	//{
	//	cout<<data1[i].a<<' '<<data1[i].b<<' '<<data1[i].c<<' '<<data1[i].d<<endl;
	//}
	dfs(2);//È¥ËÑË÷µÚx¸öÊý×Ö
	cout<<max1;
	return 0;
}