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};
use std::cell::RefCell;

fn main() {
    input! {
        n: usize, m: u64, q: usize,
        qs: [(Usize1, Usize1, u64, u64); q]
    }
    let sq = RefCell::new(Vec::new());
    let mut dfs = unsafe { std::mem::uninitialized() };
    let rec = { &dfs as *const dyn Fn() -> u64 };
    dfs = || unsafe {
        if sq.borrow().len() == n {
            qs.iter()
                .map(|&(a, b, c, d)| {
                    if sq.borrow()[b] - sq.borrow()[a] == c {
                        d
                    } else {
                        0
                    }
                })
                .sum()
        } else {
            let last = *sq.borrow().last().unwrap_or(&1);
            (last..=m)
                .map(|x| {
                    sq.borrow_mut().push(x);
                    let ret = (*rec)();
                    sq.borrow_mut().pop();
                    ret
                })
                .max()
                .unwrap()
        }
    };
    println!("{}", dfs());
}