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: Wrong Answer)

#[allow(unused_imports)]
use itertools::Itertools;
use proconio::input;
#[allow(unused_imports)]
use proconio::marker::*;

fn f(v: &[usize]) -> bool {
    v[0] == v[1] && v[1] == v[2]
}

fn main() {
    input! {
        n: usize,
        mut a: [Usize1; 3*n],
    }

    let mut t = vec![0; n];
    for &a in &a {
        t[a] += 1;
    }

    let mut res = 0;
    for i in 0..n - 1 {
        let b = &mut a[i * 3..i * 3 + 5];
        b.sort_by_key(|a| (t[*a], *a));
        if b[1] == b[2] && b[2] == b[3] {
            b.swap(0, 3);
        } else if b[2] == b[3] && b[3] == b[4] {
            b.swap(0, 3);
            b.swap(1, 4);
        }
        if f(b) {
            res += 1;
        }
        // println!("{:?}", a);

        for j in 0..3 {
            t[a[i * 3 + j]] -= 1;
        }
    }
    if f(&a[3 * n - 3..]) {
        res += 1;
    }

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