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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bit>
#include <bitset>
#include <assert.h>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <sstream>
#include <iostream>
#include <set>
#include <map>
#include <vector>
#include <math.h>
#include <stdlib.h>
#include <queue>
#include <stack>
#include <iomanip>

typedef long double ld;
typedef long long ll;

using namespace std;


/* TEMPLATE ENDS HERE */

#define MAXN (1 << 21)

vector<int> a[MAXN];
int n;

int edgeid(int i, int j) {
	if (i > j) {
		swap(i, j);
	}
	return (i - 1) * n + j - 1;
}

void addedge(int p1, int p21, int p22) {
	int ffrom = edgeid(p1, p21);
	int tto = edgeid(p1, p22);	
	//cerr << ffrom << " " << tto << endl;
	a[ffrom].push_back(tto);
}

bool ok;
int v[MAXN];
int f[MAXN];
int t;
vector<int> top;

void dfs(int cur) {
	if (v[cur]) {
		return;
	}
	v[cur] = 1;
//	t++;
	for (auto& e: a[cur]) {
		if (v[e] == 1) {
			ok = false;
			return;
		}
		dfs(e);
	}
//	t++;
//	f[cur] = t;	
	v[cur] = 2;
	top.push_back(cur);
}

int mem[MAXN];

int solve(int cur) {
	int & ret = mem[cur];
	if (ret >= 0) {
		return 0;
	}
	ret = 0;
	for (auto& e : a[cur]) {
		int nnext = solve(e);
		ret = max(nnext, ret);
	}
	ret++;
	return ret;
}

int m;

void print() {
	for (int i = 0; i < m; i++) {
		//cerr << i << ": ";
		for (auto& e : a[i]) {
			//cerr << e << " ";
		}
		//cerr << endl;
	}
}

int main() {
	cout << fixed << setprecision(12);	
	cin >> n;
	for (int i = 0; i < n; i++) {
		int cur, prev;
		for (int j = 0; j < n - 1; j++) {
			cin >> cur;		
			if (j) {
				//cerr << "here: " << i + 1 << " " << prev << " " << cur << endl;
				addedge(i + 1, prev, cur);
			}
			prev = cur;
		}
	}
	ok = true;
	m = (n * (n - 1)) / 2;
	print();
	for (int i = 0; i < m; i++) {
		dfs(i);
	}
	if (!ok) {
		cout << "-1" << endl;
		return 0;
	}
	memset(mem, -1, sizeof(mem));
	reverse(top.begin(), top.end());
	//cerr << top.size() << endl;
	int ret = 0;
	for (auto& e: top) {
		int cur = solve(e);
		//cerr << e << " " << cur << endl;
		ret = max(ret, cur);
	}
	cout << ret << endl;
	return 0;
}


ll MOD = ((ll)1e9 + 7);

struct mod {
	ll x;

	mod(ll tx = 0) {
		x = tx;
		x %= MOD;
	}
	
	mod& operator=(const ll & tx) {
		x = tx % MOD;
		return *this;
	}
	
	mod& operator=(const mod & m2) {
		x = m2.x;
		return *this;
	}

	mod operator+(const mod& m2) const {
		return mod((x + m2.x) % MOD);
	}

	mod& operator+=(const mod& m2) {
		x += m2.x;
		x %= MOD;
		return *this;
	}

	mod operator-(const mod & m2) const {
		return mod((x - m2.x + MOD) % MOD);
	}

	mod& operator-=(const mod & m2) {
		x -= m2.x;
		x = (x + MOD) % MOD;
		return *this;
	}

	mod operator*(const mod & m2) const {
		return mod((x * m2.x) % MOD);
	}

	mod& operator*=(const mod & m2) {
		x *= m2.x;
		x %= MOD;
		return *this;
	}

	mod mypow(ll k) const {
		mod ret = 1;
		mod cur = x;
		while (k) {
			if (k & 1) {
				ret *= cur;
			}
			cur *= cur;
			k >>= 1;
		}
		return ret;
	}

	mod inv() const {
		return mypow(MOD - 2);
	}

	mod operator/(const mod & m2) const {
		return (*this) * m2.inv();
	}

	mod& operator/=(const mod & m2) {
		x *= m2.inv().x;
		x %= MOD;
		return *this;
	}

	bool operator<(const mod & m2) const {
		return x < m2.x;
	}
};

ostream &operator<<(ostream & os, const mod & m) {
	return os << m.x;
}