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;
typedef long long ll;
#define rep(i, srt, end) for (long long i = (srt); i < (long long)(end); i++)
#define VL vector<ll>
#define VS vector<string>
#define VB vector<bool>
#define VP vector<pair<ll,ll>>
#define VVL vector<vector<ll>>
#define VVP vector<vector<pair<ll,ll>>>
#define PL pair<ll,ll>
#define ALL(v) (v).begin(), (v).end()
  
ll N,M,Q;
VL a,b,c,d;
ll ans=0;//初期化忘れない

void dfs(VL A){
  if((int)A.size()==(int)N+1){
    //sizeメンバ関数の戻り値はvector<型名>::size_type型
    ll now=0;//初期化忘れない
    rep(i,0,Q){
      if(A[b[i]]-A[a[i]]==c[i])now+=d[i];
    }
    ans=max(ans,now);
    return;
  }
  A.push_back(A.back());
  while(A.back()<=M){
    dfs(A);
    A.back()++;
  }
}

int main(){
  cin>>N>>M>>Q;
  a=b=c=d=VL(Q);
  rep(i,0,Q){
    cin>>a[i]>>b[i];
    cin>>c[i]>>d[i];
  }
  ans=0;
  dfs(VL (1,1));
  cout<<ans;
  return 0;
}