Problem:
Score : 500 points

Problem Statement

N players will participate in a tennis tournament. We will call them Player 1, Player 2, \ldots, Player N.

The tournament is round-robin format, and there will be N(N-1)/2 matches in total.
Is it possible to schedule these matches so that all of the following conditions are satisfied? If the answer is yes, also find the minimum number of days required.

Each player plays at most one matches in a day.

Each player i (1 \leq i \leq N) plays one match against Player A_{i, 1}, A_{i, 2}, \ldots, A_{i, N-1} in this order.

Constraints

3 \leq N \leq 1000

1 \leq A_{i, j} \leq N

A_{i, j} \neq i

A_{i, 1}, A_{i, 2}, \ldots, A_{i, N-1} are all different.

Input

Input is given from Standard Input in the following format:

N
A_{1, 1} A_{1, 2} \ldots A_{1, N-1}
A_{2, 1} A_{2, 2} \ldots A_{2, N-1}
:
A_{N, 1} A_{N, 2} \ldots A_{N, N-1}

Output

If it is possible to schedule all the matches so that all of the conditions are satisfied, print the minimum number of days required; if it is impossible, print -1.

Sample Input 1

3
2 3
1 3
1 2

Sample Output 1

3

All the conditions can be satisfied if the matches are scheduled for three days as follows:

Day 1: Player 1 vs Player 2

Day 2: Player 1 vs Player 3

Day 3: Player 2 vs Player 3

This is the minimum number of days required.

Sample Input 2

4
2 3 4
1 3 4
4 1 2
3 1 2

Sample Output 2

4

All the conditions can be satisfied if the matches are scheduled for four days as follows:

Day 1: Player 1 vs Player 2, Player 3 vs Player 4

Day 2: Player 1 vs Player 3

Day 3: Player 1 vs Player 4, Player 2 vs Player 3

Day 4: Player 2 vs Player 4

This is the minimum number of days required.

Sample Input 3

3
2 3
3 1
1 2

Sample Output 3

-1

Any scheduling of the matches violates some condition.

Solution in Rust (1.15.1) (judge verdict: Time Limit Exceeded)

use std::io::*;
use std::str::FromStr;

pub fn read<T: FromStr>() -> T {
  let stdin = stdin();
  let stdin = stdin.lock();
  let token: String = stdin
    .bytes()
    .map(|c| c.expect("failed to read char") as char)
    .skip_while(|c| c.is_whitespace())
    .take_while(|c| !c.is_whitespace())
    .collect();
  token.parse().ok().expect("failed to parse token")
}

pub fn read_n<T: FromStr>(n: usize) -> Vec<T> {
  read_n_logic::<T>(n, vec![])
}

pub fn read_n_logic<T: FromStr>(n: usize, mut a: Vec<T>) -> Vec<T> {
  match n {
    0 => a,
    _ => {
      a.push(read::<T>());
      read_n_logic(n - 1, a)
    }
  }
}

use std::cmp::{max, min};

use std::collections::BTreeMap;
use std::collections::VecDeque;

fn main() {

  let n: i64 = read();
  let mut bh: BTreeMap<i64, VecDeque<i64>> = BTreeMap::new();

  let mut ans = 0;
  for i in 0..n {
    let mut v = VecDeque::new();
    for _ in 0..n - 1 {
      let item: i64 = read();
      v.push_back(item)
    }
    bh.insert(i, v);
  }


  while true {
    ans += 1;
    let mut b = vec![];
    for i in 0..n {
      let current = bh.get_mut(&i).unwrap();
      let b_i = match current.pop_front() {
        Some(s) => s,
        _ => -1,
      };
      if b_i <= b.len() as i64 && b_i != -1 && b[(b_i - 1) as usize] == i + 1 {
        b.push(0);
        b[(b_i - 1) as usize] = 0;
      } else {
        b.push(b_i)
      }
    }

    // TODO 全部-1で終了
    if b.iter().all(|x| *x == -1) {
      break;
    }

    // TODO 0が一つもなかったら-1になる
    if b.iter().all(|x| *x != 0) {
      ans = -1;
      break;
    }

    for (i, v) in b.iter().enumerate() {
      let i_int = i as i64;
      if *v != 0 && *v != -1 {
        let current = bh.get_mut(&i_int).unwrap();
        current.push_front(*v);
      }
    }
  }

  if ans == -1 {
    println!("{}", ans);
  } else {
    println!("{}", ans - 1);
  }

}