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)

fn main() {
    proconio::input! {
        n: usize,
        a1: proconio::marker::Usize1,
        a2: proconio::marker::Usize1,
        a: [(proconio::marker::Usize1, proconio::marker::Usize1, proconio::marker::Usize1); n-1],
        an: proconio::marker::Usize1
    };
    let mut dp = vec![vec![-1; n]; n];
    let mut all = 0;
    dp[a1][a2] = 0;
    for &p in &a {
        if p.0 == p.1 && p.1 == p.2 {
            all += 1;
            continue;
        }
        let mut update = Vec::new();
        update.push((p.1, p.2, if dp[p.0][p.0] != -1 { dp[p.0][p.0] + 1 } else { 0 }));
        update.push((p.2, p.0, if dp[p.1][p.1] != -1 { dp[p.1][p.1] + 1 } else { 0 }));
        update.push((p.0, p.1, if dp[p.2][p.2] != -1 { dp[p.2][p.2] + 1 } else { 0 }));
        for i in 0..n {
            if dp[i][p.1] != -1 {
                update.push((i, p.0, dp[i][p.1] + if p.1 == p.2 { 1 } else { 0 }));
            }
            if dp[i][p.2] != -1 {
                update.push((i, p.1, dp[i][p.2] + if p.2 == p.0 { 1 } else { 0 }));
            }
            if dp[i][p.0] != -1 {
                update.push((i, p.2, dp[i][p.0] + if p.0 == p.1 { 1 } else { 0 }));
            }
        }
        for &u in &update {
            dp[u.0][u.1] = dp[u.0][u.1].max(u.2);
            dp[u.1][u.0] = dp[u.1][u.0].max(u.2);
        }
    }
    let mut res = 0;
    for i in 0..n {
        for j in 0..n {
            res = res.max(dp[i][j] + all + if i == j && i == an { 1 } else { 0 });
        }
    }
    println!("{}", res);
}