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)

fun main(args:Array<String>){
    val l = readLine()!!.split(" ").map {it.toInt()}
    val n = l[0]
    val m = l[1]
    val q = l[2]
    val a = Array(q){0}
    val b = Array(q){0}
    val c = Array(q){0}
    val d = Array(q){0}
    (0..q-1).map {
        val ll = readLine()!!.split(" ").map {it.toInt()}
        a[it] = ll[0]-1
        b[it] = ll[1]-1
        c[it] = ll[2]
        d[it] = ll[3]
    }
    var result = 0
    fun dfs(arg:Pair<Int,Int>,array:Array<Int>){
        val current = arg.first
        val max = arg.second
        //実装部分
        var sum = 0
        (0..q-1).map{
            if(array[b[it]]-array[a[it]]==c[it]){
                sum += d[it]
            }
        }
        if(sum>result)result=sum
        if(current>0){
            val copy = array.copyOf()
            copy[current-1]++
            if(copy[current-1]<=copy[current]) {
                dfs(Pair(current - 1, max), copy)
            }
        }
        if(array[current]<max){
            val copy = array.copyOf()
            copy[current]++
            dfs(Pair(current,max),copy)
        }
    }
    dfs(Pair(n-1,m),Array(n){1})
    println(result)
}