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: Wrong Answer)

use proconio::input;

fn calc_forward(v: &mut Vec<usize>, m: usize) {
    for i in (1..v.len()).rev() {
        if v[i] > m {
            v[i-1] += 1;
            v[i] -= m;
        }
    }
}
fn calc_backward(v: &mut Vec<usize>) {
    for i in 0..(v.len() - 1) {
        if v[i] > v[i+1] {
            v[i+1] = v[i];
        }
    }
}

fn main() {
    input! {
        n: usize,
        m: usize,
        q: usize,
        abcd: [[usize; 4]; q],
    }
    let mut result = 0;
    let mut v = vec![0; n];
    while v[0] <= 10 {
        let mut r = 0;
        for i in 0..q {
            let a = abcd[i][0] - 1;
            let b = abcd[i][1] - 1;
            let c = abcd[i][2];
            let d = abcd[i][3];

            if v[b] - v[a] == c {
                r += d;
            }
        }
        result = result.max(r);

        v[n - 1] += 1;
        calc_forward(&mut v, m);
        calc_backward(&mut v);
    }

    println!("{}", result);
}