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>
#pragma GCC target("avx")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
using i64 = int64_t;
using f80 = __float80;
static const int dy[4] = {0,1,0,-1}, dx[4] = {1,0,-1,0};
constexpr i64 INF = 1'010'000'000'000'000'017LL;
constexpr i64 MOD = 1'000'000'007LL;
#define _overload3(_1,_2,_3,name,...) name
#define repi(i,start,end) for(i64 i=(start);i<(end);++i)
#define _rep(i,n) repi(i,0,n)
#define rep(...) _overload3(__VA_ARGS__,repi,_rep,)(__VA_ARGS__)
#define rrep(i,start,end) for(i64 i=(start);i>end;--i)
#define rng(a) a.begin(),a.end()
#define rrng(a) a.rbegin(),a.rend()
#define isin(x,l,r) ((l) <= (x) && (x) < (r))
#define show(x) cout<<#x<<" = "<<x<<endl;
#define PQ(T) priority_queue<T,v(T),greater<T> >
#define v(T) vector<T>
#define vv(T) v(v(T))
typedef v(i64) vi;
typedef v(vi) vvi;
template<class T> inline bool chmax(T& a,T b){if(a<b){a=b;return true;}return false;}
template<class T> inline bool chmin(T& a,T b){if(a>b){a=b;return true;}return false;}
template<typename T> ostream &operator<<(ostream &os, const vector< T > &v){
  rep(i,v.size()) os << v[i] << (i + 1 != v.size() ? " " : "");
  return os;
}
template<typename T> istream &operator>>(istream &is, vector< T > &v){
  for(T &in : v) is >> in;
  return is;
}

vvi vA;

void bfn(vi A, i64 N, i64 M) {
  if(A.size() == N) {
    vA.emplace_back(A);
    return;
  }
  rep(i, A[A.size() - 1], M + 1){
    vi nA;
    for(auto &p : A) nA.emplace_back(p);
    nA.emplace_back(i);
    bfn(nA, N, M);
  }
  return;
}

void Main()
{
  i64 N, M, Q;
  cin >> N >> M >> Q;
  vvi ABCD(Q, vi(4));
  cin >> ABCD;

  vi A;
  A.emplace_back(1);

  bfn(A, N, M);

  i64 ans = 0;

  for(auto &p : vA){
    i64 ans_tmp = 0;
    rep(i,Q){
      if(p[ABCD[i][1]-1] - p[ABCD[i][0]-1] == ABCD[i][2]) ans_tmp += ABCD[i][3];
    }
    chmax(ans, ans_tmp);
  }

  cout << ans << "\n";




}

int main()
{
  cin.tie(nullptr);
  ios_base::sync_with_stdio(false);
  cout << fixed << setprecision(15);
  Main();
}