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 Kotlin (1.3.71) (judge verdict: Wrong Answer)

import java.util.*


fun main() {
    val sc = Scanner(System.`in`)
    val n = sc.nextInt()
    val m = sc.nextInt()
    val q = sc.nextInt()

    data class ABCD(val a: Int, val b: Int, val c: Int, val d: Int)

    val tuples = Array(q) { ABCD(sc.nextInt(), sc.nextInt(), sc.nextInt(), sc.nextInt()) }

    var maxSums = 0L
    for (listA in getAs(n, m)) {
        var sum = 0L
        for ((a, b, c, d) in tuples) {
            if (listA[b-1] - listA[a-1] == c)
                sum += d
        }
        maxSums = maxOf(maxSums, sum)
    }

    println(maxSums)
}

fun getAs(n: Int, m: Int) = sequence<IntArray> {
    fun a(prev: Int): IntRange {
        return (prev) .. (m)
    }
    fun b(ary: IntArray, dims: Int): Sequence<IntArray> = sequence {
        if (dims == 0) yield(ary)
        else {
            for (v in a(ary.getOrNull(ary.size - dims - 1) ?: 0)) {
                ary[ary.size - dims] = v
                yieldAll(b(ary, dims - 1))
            }
        }
    }
    yieldAll(b(IntArray(n), n))
}