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 <ext/pb_ds/assoc_container.hpp> 
#include <ext/pb_ds/tree_policy.hpp> 
#define ll long long int
#define MOD 1000000007
#define oo 1000000000000000000
#define forr(i,n) for(ll i=0;i<n;i++)
#define fastio ios_base::sync_with_stdio(false); cin.tie(0) ; cout.tie(0);
#define all(x) x.begin(),x.end()
#define eb emplace_back
#define pb push_back
#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update> 
 
using namespace __gnu_pbds; 
using namespace std;

ll valueOfIndex(ordered_set&s , ll i){ return *(s.find_by_order(i)); }
ll indexOfValue(ordered_set&s , ll x){ return s.order_of_key(x); }
ll add(ll a, ll b,ll p=MOD) { a%=p; b%=p; return (a+b + p)%p;}
ll mul(ll a, ll b,ll p=MOD) { a%=p; b%=p; return (a*b + p)%p;} // __int128
ll power(ll x,ll n,ll p=MOD){ if(x==0) return 0; if(n==0 || x==1) return 1LL;
    ll r = (power(x,n/2,p))%p; if(n&1) return mul(mul(r,r,p) , x,p); else return mul(r,r,p);
}
ll inv(ll x){return power(x,MOD-2);}

vector<int> v(1000,0);

pair<pair<int,int> , pair<int,int> > arr[1000]; // a b c d
int n,m,q;

int ans = INT_MIN;

int score(){
    if(v.size()<q+1) return 0;
    int cnt=0;
    for(int i=0;i<q;i++){
        if( v[arr[i].first.second] - v[arr[i].first.first] == arr[i].second.first ){
            cnt+= arr[i].second.second;
        }
    }
    return cnt;
}

void rec(int j){
    if(j==n+1){
        ans = max(ans , score());
        return;
    }
    for(int i=1;i<=m;i++){
        v[j] = i ;
        rec(j+1);
    }
}


void __sol(){
    cin >> n >> m >> q;
    forr(i,q) cin >> arr[i].first.first >> arr[i].first.second >> arr[i].second.first >> arr[i].second.second;
    rec(0);
    cout << ans;

}


int main(){
    fastio
    ll tc=1; //cin >> tc;
    while(tc--) __sol();
    return 0;
}