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: Wrong Answer)

#[allow(unused_imports)]
use std::cmp;
use std::io::Write;

// Input macro from https://qiita.com/tanakh/items/0ba42c7ca36cd29d0ac8
macro_rules! input {
  (source = $s:expr, $($r:tt)*) => {
    let mut iter = $s.split_whitespace();
    input_inner!{iter, $($r)*}
  };
  ($($r:tt)*) => {
    let s = {
      use std::io::Read;
      let mut s = String::new();
      std::io::stdin().read_to_string(&mut s).unwrap();
      s
    };
    let mut iter = s.split_whitespace();
    input_inner!{iter, $($r)*}
  };
}

macro_rules! input_inner {
  ($iter:expr) => {};
  ($iter:expr, ) => {};

  ($iter:expr, $var:ident : $t:tt $($r:tt)*) => {
    let $var = read_value!($iter, $t);
    input_inner!{$iter $($r)*}
  };
}

macro_rules! read_value {
  ($iter:expr, ( $($t:tt),* )) => {
    ( $(read_value!($iter, $t)),* )
  };

  ($iter:expr, [ $t:tt ; $len:expr ]) => {
    (0..$len).map(|_| read_value!($iter, $t)).collect::<Vec<_>>()
  };

  ($iter:expr, chars) => {
    read_value!($iter, String).chars().collect::<Vec<char>>()
  };

  ($iter:expr, usize1) => {
    read_value!($iter, usize) - 1
  };

  ($iter:expr, $t:ty) => {
    $iter.next().unwrap().parse::<$t>().expect("Parse error")
  };
}

type Index = (usize, usize);

fn get_index(i: usize, j: usize) -> Index {
  if i < j { (i, j) } else { (j, i) }
}

use std::collections::{BTreeMap, BTreeSet};

fn longest_path(v: Index, adj: &BTreeMap<Index, Vec<Index>>, len: &mut BTreeMap<Index, usize>, active: &mut BTreeSet<Index>) -> Option<usize> {
  if let Some(l) = len.get(&v) {
    return Some(*l);
  }
  if active.contains(&v) {
    return None;
  }
  let mut l = 0;
  active.insert(v);
  if let Some(adj_v) = adj.get(&v) {
    for &next in adj_v {
      if let Some(nl) = longest_path(next, adj, len, active) {
        l = cmp::max(l, nl + 1);
      } else {
        return None;
      }
    }
  }
  len.insert(v, l);
  Some(l)
}

fn main() {
  let out = std::io::stdout();
  let mut out = std::io::BufWriter::new(out.lock());
  macro_rules! puts {
    ($($format:tt)*) => (write!(out,$($format)*).unwrap());
  }
  input! {
    n: usize,
    a: [[usize1; n-1]; n],
  }
  let mut adj = BTreeMap::new();
  for i in 0 .. n {
    for j in 0 .. n - 2 {
      let index = get_index(i, a[i][j]);
      let next = adj.entry(index).or_insert(vec![]);
      next.push(get_index(i, a[i][j+1]))
    }
  }
  let mut len = BTreeMap::new();
  let mut ans = 0;
  let mut active = BTreeSet::new();
  for &idx in adj.keys() {
    if let Some(l) = longest_path(idx, &adj, &mut len, &mut active) {
      ans = cmp::max(ans, l as i64 + 1);
    } else {
      ans = -1;
    }
  }

  puts!("{}\n", ans);
}