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 Ruby (2.7.1) (judge verdict: Accepted)

N, M, _ = STDIN.gets.split(' ').map(&:to_i)
Q = STDIN.read.split("\n").map {|line| line.split(' ').map(&:to_i) }

A = (1..M).to_a

def cal(arr)
  point = 0

  Q.each do |(a, b, c, d)|
    if arr[b-1] - arr[a-1] == c
      point += d
    end
  end

  point
end

max = 0

case N
when 2
  1.upto(M) do |a1|
    a1.upto(M) do |a2|
      v = cal([a1, a2])
      max = [v, max].max
    end
  end
when 3
  1.upto(M) do |a1|
    a1.upto(M) do |a2|
      a2.upto(M) do |a3|
        v = cal([a1, a2, a3])
        max = [v, max].max
      end
    end
  end
when 4
  1.upto(M) do |a1|
    a1.upto(M) do |a2|
      a2.upto(M) do |a3|
        a3.upto(M) do |a4|
          v = cal([a1, a2, a3, a4])
          max = [v, max].max
        end
      end
    end
  end
when 5
  1.upto(M) do |a1|
    a1.upto(M) do |a2|
      a2.upto(M) do |a3|
        a3.upto(M) do |a4|
          a4.upto(M) do |a5|
            v = cal([a1, a2, a3, a4, a5])
            max = [v, max].max
          end
        end
      end
    end
  end
when 6
  1.upto(M) do |a1|
    a1.upto(M) do |a2|
      a2.upto(M) do |a3|
        a3.upto(M) do |a4|
          a4.upto(M) do |a5|
            a5.upto(M) do |a6|
              v = cal([a1, a2, a3, a4, a5, a6])
              max = [v, max].max
            end
          end
        end
      end
    end
  end
when 7
  1.upto(M) do |a1|
    a1.upto(M) do |a2|
      a2.upto(M) do |a3|
        a3.upto(M) do |a4|
          a4.upto(M) do |a5|
            a5.upto(M) do |a6|
              a6.upto(M) do |a7|
                v = cal([a1, a2, a3, a4, a5, a6, a7])
                max = [v, max].max
              end
            end
          end
        end
      end
    end
  end
when 8
  1.upto(M) do |a1|
    a1.upto(M) do |a2|
      a2.upto(M) do |a3|
        a3.upto(M) do |a4|
          a4.upto(M) do |a5|
            a5.upto(M) do |a6|
              a6.upto(M) do |a7|
                a7.upto(M) do |a8|
                  v = cal([a1, a2, a3, a4, a5, a6, a7, a8])
                  max = [v, max].max
                end
              end
            end
          end
        end
      end
    end
  end
when 9
  1.upto(M) do |a1|
    a1.upto(M) do |a2|
      a2.upto(M) do |a3|
        a3.upto(M) do |a4|
          a4.upto(M) do |a5|
            a5.upto(M) do |a6|
              a6.upto(M) do |a7|
                a7.upto(M) do |a8|
                  a8.upto(M) do |a9|
                    v = cal([a1, a2, a3, a4, a5, a6, a7, a8, a9])
                    max = [v, max].max
                  end
                end
              end
            end
          end
        end
      end
    end
  end
when 10
  1.upto(M) do |a1|
    a1.upto(M) do |a2|
      a2.upto(M) do |a3|
        a3.upto(M) do |a4|
          a4.upto(M) do |a5|
            a5.upto(M) do |a6|
              a6.upto(M) do |a7|
                a7.upto(M) do |a8|
                  a8.upto(M) do |a9|
                    a9.upto(M) do |a10|
                      v = cal([a1, a2, a3, a4, a5, a6, a7, a8, a9, a10])
                      max = [v, max].max
                    end
                  end
                end
              end
            end
          end
        end
      end
    end
  end
end

print max