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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define whlie while
#define pb push_back
#define eb emplace_back
#define fi first
#define se second
#define rep(i,N) for(int i = 0; i < (N); i++)
#define repr(i,N) for(int i = (N) - 1; i >= 0; i--)
#define rep1(i,N) for(int i = 1; i <= (N) ; i++)
#define repr1(i,N) for(int i = (N) ; i > 0 ; i--)
#define each(x,v) for(auto& x : v)
#define all(v) (v).begin(),(v).end()
#define sz(v) ((int)(v).size())
#define vrep(v,it) for(auto it = v.begin(); it != v.end(); it++)
#define vrepr(v,it) for(auto it = v.rbegin(); it != v.rend(); it++)
#define ini(...) int __VA_ARGS__; in(__VA_ARGS__)
#define inl(...) ll __VA_ARGS__; in(__VA_ARGS__)
#define ins(...) string __VA_ARGS__; in(__VA_ARGS__)
#define rand rng
using namespace std; void solve();
using ll = long long;
using vi = vector<int>;
using vl = vector<ll>;
using vvi = vector< vector<int> >;
constexpr int inf = 1001001001;
constexpr ll infLL = (1LL << 61) - 1;

struct IoSetupNya {IoSetupNya() { cin.tie(nullptr); ios::sync_with_stdio(false); cout << fixed << setprecision(15); cerr << fixed << setprecision(7);} } iosetupnya;
template<typename T, typename U> inline bool amin(T &x, U y) { return (y < x) ? (x = y, true) : false; }
template<typename T, typename U> inline bool amax(T &x, U y) { return (x < y) ? (x = y, true) : false; }
template<typename T, typename U> ostream& operator <<(ostream& os, const pair<T, U> &p) { os << p.first << " " << p.second; return os; }
template<typename T, typename U> istream& operator >>(istream& is, pair<T, U> &p) { is >> p.first >> p.second; return is; }
template<typename T> ostream& operator <<(ostream& os, const vector<T> &v) { int s = (int)v.size(); rep(i,s) os << (i ? " " : "") << v[i]; return os; }
template<typename T> istream& operator >>(istream& is, vector<T> &v) { for(auto &x : v) is >> x; return is; }
void in(){} template <typename T,class... U> void in(T &t,U &...u){ cin >> t; in(u...);}
void out(){cout << "\n";} template <typename T,class... U> void out(const T &t,const U &...u){ cout << t; if(sizeof...(u)) cout << " "; out(u...);}
template<typename T>void die(T x){out(x); exit(0);}
#ifdef NyaanDebug
  #include "NyaanDebug.h"
  #define trc(...) do { cerr << #__VA_ARGS__ << " = "; dbg_out(__VA_ARGS__);} while(0)
  #define trca(v,N) do { cerr << #v << " = "; array_out(v , N);cout << endl;} while(0)
  int main(){solve_dbg();}
#else
  #define trc(...)
  #define trca(...)
  int main(){solve();}
#endif

using P = pair<int,int>;
constexpr int MOD = 998244853;
mt19937_64 rng(58);
template<typename T>inline T ceil(T a,T b){return (a + b - 1) / b;}
/////////

// トポロジカルソート 任意のグラフに使用可能(たぶん) O(E+V)
// DAGであるときはtrueを、閉路が存在するときfalseを返す
template<typename T>
bool TopologicalSort(vector<int> &v , T &g){
  int N = g.size();
  vector<int> marked(N , 0) , temp(N , 0);
  v.clear(); v.reserve(N);
  // 訪問して帰りがけ順にvに追加
  auto visit = [&](auto f, int i) -> bool{
    if(temp[i] == 1) return false;
    if(marked[i] == 0){
      temp[i] = 1;
      for(auto &e : g[i]){
        if(f(f , e) == false) return false;
      }
      marked[i] = 1;
      v.push_back(i);
      temp[i] = 0;
    }
    return true;
  };

  for(int i = 0 ; i < N ; i++){
    if(marked[i] == 0) {
      if(visit(visit , i) == false) return false;
    }
  }
  reverse( v.begin() , v.end() );
  return true;
}

vvi g;

void solve(){
  ini(N);
  vvi a(N , vi(N - 1));
  rep(i , N) in(a[i]);
  rep(i,N) rep(j,N - 1) a[i][j]--;
  rep(i , N) rep(j , N - 1){
    a[i][j] = min(i , a[i][j]) * N + max(i , a[i][j]);
  }
  rep(i,N) trc(a[i]);
  g.resize(N * N);
  rep(i , N) rep(j , N - 2){
    g[a[i][j]].pb(a[i][j + 1]);
  }
  trc("HOGE");

  vi topo;
  if(TopologicalSort(topo , g) == false) die(-1);
  {  
    vi topo2; topo2.reserve(N * (N - 1) / 2);
    each(x,topo){
      if(x / N < x % N) topo2.pb(x);
    }
    topo.swap(topo2);
  }
  trc(topo);

  vi inv(N*N , -1);
  rep(i , sz(topo)) inv[topo[i]] = i;
  trc(inv);

  vi ans(sz(topo) , 1);
  int cnt = 0;
  each(i , topo){
    if(g[i].empty()) continue;
    each(x , g[i]){
      amax( ans[inv[x]] , ans[inv[i]] + 1 );
      amax( cnt , ans[inv[x]]);
    }
  }
  trc(ans);
  out(cnt);


}