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::*;
use proconio::marker::Usize1;

fn dfs(n: usize, m: usize, a: &mut [usize], g: &[Vec<(usize, usize, u32)>], score: u32) -> u32 {
    if n == 0 {
        return score;
    }
    let mut ans = 0;
    for i in 0..m {
        a[n - 1] = i;
        let mut s = score;
        for &(b, c, d) in g[n - 1].iter() {
            if a[b] - a[n - 1] == c {
                s += d;
            }
        }
        ans = std::cmp::max(ans, dfs(n - 1, i + 1, a, g, s));
    }
    ans
}

fn run() {
    input! {
        n: usize,
        m: usize,
        q: usize,
        e: [(Usize1, Usize1, usize, u32); q],
    }
    let mut g = vec![vec![]; n];
    for (a, b, c, d) in e {
        g[a].push((b, c, d));
    }
    let mut val = vec![0; n];
    val[n - 1] = m - 1;
    let ans = dfs(n - 1, m, &mut val, &g, 0);
    println!("{}", ans);
}

fn main() {
    run();
}