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 Rust (1.42.0) (judge verdict: Accepted)

use proconio::{input, marker::Usize1};

fn main() {
    input! {
        n: usize, m: u64, q: usize,
        qs: [(Usize1, Usize1, u64, u64); q]
    }
    let mut g = G {
        n,
        m,
        qs,
        sq: Vec::new(),
    };
    println!("{}", g.dfs())
}

struct G {
    n: usize,
    m: u64,
    qs: Vec<(usize, usize, u64, u64)>,
    sq: Vec<u64>,
}

impl G {
    fn dfs(&mut self) -> u64 {
        if self.sq.len() == self.n {
            let ev = |&(a, b, c, d)| if self.sq[b] - self.sq[a] == c { d } else { 0 };
            self.qs.iter().map(ev).sum()
        } else {
            let last = *self.sq.last().unwrap_or(&1);
            (last..=self.m)
                .map(|x| {
                    self.sq.push(x);
                    let ret = self.dfs();
                    self.sq.pop();
                    ret
                })
                .max()
                .unwrap()
        }
    }
}