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>

#define REP(i,s,n) for(int i=s;i<n;++i)
#define rep(i,n) REP(i,0,n)
#define fst first
#define snd second
#define pb push_back
#define ALL(x) x.begin(),x.end()
#define EPS (1e-9)
#define equals(a,b) (fabs((a)-(b))<EPS)

using namespace std;

bool LT(double a,double b) { return !equals(a,b) && a < b; }
bool LTE(double a,double b) { return equals(a,b) || a < b; }

const string YES = "";
const string NO  = "";

typedef long long ll;
typedef pair<int,int> ii;

#define MAX 51
int N,M,Q,a[MAX], b[MAX], c[MAX], d[MAX];

int ans;
void dfs(deque<int> &deq, int cur) {
  if( cur >= N ) {
    int sum = 0;
    rep(i,Q) {
      if( deq[b[i]] - deq[a[i]] == c[i] ) {
	sum += d[i];
      }
    }
    ans = max(sum,ans);
    return;
  }
  int sp = 1;
  if( deq.size() ) sp = deq.back();
  REP(i,sp,M+1) {
    deq.pb(i);
    dfs(deq,cur+1);
    deq.pop_back();
  }
}

int main() {
  cin >> N >> M >> Q;
  rep(i,Q) {
    cin >> a[i] >> b[i] >> c[i] >> d[i];
    --a[i], --b[i];
  }
  deque<int> deq;
  dfs(deq,0);
  cout << ans << endl;
  return 0;
}