Problem:
Score : 600 points

Problem Statement

We have 3N cards arranged in a row from left to right, where each card has an integer between 1 and N (inclusive) written on it. The integer written on the i-th card from the left is A_i.

You will do the following operation N-1 times:

Rearrange the five leftmost cards in any order you like, then remove the three leftmost cards. If the integers written on those three cards are all equal, you gain 1 point.

After these N-1 operations, if the integers written on the remaining three cards are all equal, you will gain 1 additional point.

Find the maximum number of points you can gain.

Constraints

1 \leq N \leq 2000

1 \leq A_i \leq N

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 \cdots A_{3N}

Output

Print the maximum number of points you can gain.

Sample Input 1

2
1 2 1 2 2 1

Sample Output 1

2

Let us rearrange the five leftmost cards so that the integers written on the six cards will be 2\ 2\ 2\ 1\ 1\ 1 from left to right.

Then, remove the three leftmost cards, all of which have the same integer 2, gaining 1 point.

Now, the integers written on the remaining cards are 1\ 1\ 1.

Since these three cards have the same integer 1, we gain 1 more point.

In this way, we can gain 2 points - which is the maximum possible.

Sample Input 2

3
1 1 2 2 3 3 3 2 1

Sample Output 2

1

Sample Input 3

3
1 1 2 2 2 3 3 3 1

Sample Output 3

3

Solution in Rust (1.42.0) (judge verdict: Time Limit Exceeded)

#[allow(unused_imports)]
use std::{cmp::Reverse, collections::*};

#[allow(unused_imports)]
use proconio::{
    fastout, input,
    marker::{Chars, Usize1},
};

#[allow(unused_imports)]
use itertools::Itertools;

#[allow(unused_imports)]
use num::Integer;

#[fastout]
pub fn main() {
    input! {
        n: usize,
        a: [usize; 3 * n],
    }

    let mut q = VecDeque::new();
    q.push_back((a, 0, 0));
    let mut max = 0;
    while let Some((a, count, point)) = q.pop_front() {
        if count == n - 1 {
            let point = if a[0] == a[1] && a[0] == a[2] {
                point + 1
            } else {
                point
            };
            if max < point {
                max = point;
            }
            continue;
        }

        for i in 0..3 {
            for j in (i + 1)..4 {
                for k in (j + 1)..5 {
                    let point = if a[i] == a[j] && a[i] == a[k] {
                        point + 1
                    } else {
                        point
                    };
                    q.push_back((
                        a.iter()
                            .enumerate()
                            .filter(|&(idx, _)| idx != i && idx != j && idx != k)
                            .map(|(_, e)| *e)
                            .collect(),
                        count + 1,
                        point,
                    ));
                }
            }
        }
    }
    println!("{}", max);
}