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 Scala (2.13.1) (judge verdict: Accepted)

object Main {
  def main(args: Array[String]): Unit = {
    val sc = new java.util.Scanner(System.in)
    val n = sc.nextInt
    val m = sc.nextInt
    val q = sc.nextInt
    val a = Array.ofDim[Int](q)
    val b = Array.ofDim[Int](q)
    val c = Array.ofDim[Int](q)
    val d = Array.ofDim[Int](q)
    (0 until q).foreach { i =>
      a(i) = sc.nextInt
      b(i) = sc.nextInt
      c(i) = sc.nextInt
      d(i) = sc.nextInt
    }

    println(new QC(n, m, q, a, b, c, d).solve)
  }
}

class QC(n: Int, m: Int, q: Int, a: Array[Int], b: Array[Int], c: Array[Int], d: Array[Int]) {
  def solve: Int = {
    val array = Array.fill[Int](n)(1)
    var ans = 0
    var break = false

    def countup(index: Int, indexUp: Boolean): Unit = {
      if (array(index) < m) {
        val next = array(index) + 1
        if (indexUp) {
          (index until n).foreach(array(_) = next)
        } else {
          array(index) = next
        }
      } else {
        countup(index - 1, true)
      }
    }

    while (!break) {
      ans = Math.max(ans, calc(array))
      break = array(0) == m
      if(!break) {
        countup(n - 1, false)
      }
    }
    ans
  }

  private def calc(array: Array[Int]) = {
    (0 until q).foldLeft(0) { (acc, cur) =>
      if (array(b(cur) - 1) - array(a(cur) - 1) equals c(cur)) {
        acc + d(cur)
      } else {
        acc
      }
    }
  }
}