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)

#![allow(unused_imports)]
#![allow(clippy::many_single_char_names)]
#![allow(clippy::suspicious_arithmetic_impl)]

// using Rust(1.42.0)
// My Library Repo is at "https://github.com/Loptall/prcn_lib"
// check available crates on AtCoder at "https://atcoder.jp/contests/language-test-202001"

/*
    AtCoder Language Update 2020204 お疲れ様でした! 有難いです
*/

// Dep crates are...
use std::cmp::{max, min, Ordering};
use std::collections::{BTreeMap, BTreeSet, HashMap, HashSet, VecDeque};
use std::mem::swap;
use std::ops::*;
use std::{i128, i16, i32, i64, i8, isize, u128, u16, u32, u64, u8, usize};

use itertools::*;
use itertools_num::*;
use lazy_static::lazy_static;
use maplit::{btreemap, btreeset, hashmap, hashset};
use num_bigint::{BigInt, BigUint};
use num_integer::{binomial, gcd, lcm, Integer};
use permutohedron::Heap;
use proconio::{fastout, input, is_stdin_empty, marker::*};
use rand::random;

// code...
fn main() {
    input! {
        n: usize, m: usize, q: usize,
        rq: [(Usize1, Usize1, usize, usize); q]
    }

    let mut ans = 0usize;

    for a1 in 1..=m {
        for a2 in 1..=m {
            if a2 < a1 {
                continue;
            }
            for a3 in 1..=m {
                if a3 < a2 {
                    continue;
                }
                for a4 in 1..=m {
                    if a4 < a3 {
                        continue;
                    }
                    for a5 in 1..=m {
                        if a5 < a4 {
                            continue;
                        }
                        for a6 in 1..=m {
                            if a6 < a5 {
                                continue;
                            }
                            for a7 in 1..=m {
                                if a7 < a6 {
                                    continue;
                                }
                                for a8 in 1..=m {
                                    if a8 < a7 {
                                        continue;
                                    }
                                    for a9 in 1..=m {
                                        if a9 < a8 {
                                            continue;
                                        }
                                        for a10 in 1..=m {
                                            if a10 < a9 {
                                                continue;
                                            }
                                            let v = [a1, a2, a3, a4, a5, a6, a7, a8, a9, a10];
                                            let mut count = 0usize;
                                            for (a, b, c, d) in rq.iter() {
                                                if v[*b] - v[*a] == *c {
                                                    count += d
                                                }
                                                ans = max(ans, count);
                                            }
                                        }
                                    }
                                }
                            }
                        }
                    }
                }
            }
        }
    }

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