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 <iostream>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <vector>
#include <set>
#include <queue>
#include <map>
#include <string>

#define rep(i, a, b) for ( int i = (a); i < (b); i++ )
#define per(i, a, b) for ( int i = (b)-1; i >= (a); i--)
#define pb push_back
#define mp make_pair
#define bg begin()
#define en end()
#define all(v) (v).begin(), (v).end()
#define sz(v) (int)(v).size()


using namespace std;

typedef long long ll;

static const long long MOD = 1000000007;
static const long long INF = 1<<31;

int a[55], b[55], c[55], d[55];
int x[15];
int n, m, q;

int dfs(int p) {
  int res = 0;
  
  if (p==n+1) {
    /*
    rep(i, 1, n+1) {
      printf("%d", x[i]);
    }
    printf("\n");
    */
    rep (i, 0, q) if (x[b[i]] - x[a[i]] == c[i]) res += d[i];
    return res;
  }
  
  for (int i=x[p-1]; i<=m; i++) {
    x[p] = i;
    res = max(res, dfs(p+1));
  }
  return res;

}

int main(void) {
  cin >> n >> m >> q;
  rep(i, 0, q) {
    scanf("%d %d %d %d", &a[i], &b[i], &c[i], &d[i]);
  }
  
  x[1] = 1;
  int ans = dfs(2);
    
    cout << ans << "\n";

    return 0;
}