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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define countof(array) (sizeof(array) / sizeof(array[0]))
#define rep(i,n) for(int i = 0; i < (n); ++i)
#define rrep(i,n) for(int i = (n)-1; i >= 0; --i)
#define srep(i,n,s) for(int i = s; i < (n); ++i)
#define rsrep(i,n,s) for(int i = (n)-1; i >= s; --i)
#define all(a) (a).begin(),(a).end()
#define rall(a) (a).rbegin(),(a).rend()
#define aall(a) (a), (a)+countof(a)//for array sorting
#define raall(a) (a), (a)+countof(a), greater<>()
#define show(x) cout<<#x<<" = "<<x<<endl;
#define vfind(v, a) find(all(v), a) != v.end()
#define yn(f) { if (f) puts("YES"); else puts("NO"); }
#define yns(f) { if (f) puts("Yes"); else puts("No"); }
#define show_ary(...) { cout<<#__VA_ARGS__<<" = "; for (const auto& x : (__VA_ARGS__)) { cout<<x<<" "; } cout<<endl; }
#define show_pair(...) cout<<#__VA_ARGS__<<" = "<<endl; for (const auto& x : (__VA_ARGS__)) { cout<<"  "<<x.fi<<" : "<<x.se<<endl; }
#define out_ary(...) { for (const auto& x : (__VA_ARGS__)) { cout<<x<<" "; } cout<<endl; }
#define argmax(v) distance((v).begin(), max_element(all(v)))
#define argmin(v) distance((v).begin(), min_element(all(v)))
#define vmax(v) *max_element(all(v))
#define vmin(v) *min_element(all(v))
typedef long long int ll;
typedef pair<int, int> P;
typedef vector<P> vpair;
typedef vector<int> vint;
typedef vector<ll> vll;
typedef vector<double> vdouble;
typedef vector<string> vstr;
typedef vector<bool> vbool;
typedef vector<vint> vvint;
typedef vector<vll> vvll;
typedef vector<vstr> vvstr;
typedef vector<vbool> vvbool;
const ll LINF = 2000000000000000000ll;
const int INF = 1000000100;
const ll MOD = 1e9+7;

int main() {
    int n;
    cin >> n;
    deque<P> m;
    vvint match(n, vint(n-1));
    rep(i, n) {
        rep(j, n-1) {
            cin >> match[i][j]; 
        }
    }
    rep(j, n-1) {
        rep(i, n) {
            if (i+1 < match[i][j]) m.emplace_back(i+1, match[i][j]);
            else m.emplace_back(match[i][j], i+1);
        }
    }
    int p_a, p_b;
    int erased = 0;
    vint used;
    int ans = 0;
    while(m.size() > erased) {
        int p = erased;
        rep(i, m.size()) {
            if (m[i].fi == 0) continue;
            p_a = m[i].fi;
            p_b = m[i].se;
            bool ok = false;
            rep(j, m.size()) {
                if (i == j) continue;
                if (m[j].fi == 0) continue;
                if (m[j].fi == p_a && m[j].se == p_b) {
                    if (vfind(used, p_a) || vfind(used, p_b)){
                        used = {};
                        ans++;
                    }
                    used.push_back(p_a);
                    used.push_back(p_b);
                    ok = true;
                    m[i].fi = 0;
                    m[j].fi = 0;
                    m[i].se = 0;
                    m[j].se = 0;
                    erased += 2;
                    break;
                }
                if (m[j].fi == p_b || m[j].se == p_b) {
                    break;
                }
            }
            if (ok) break;
            if (i == m.size()-1) {
                puts("-1");
                return 0;
            }
        }
        //if (p == erased) break;
    }
    ans++;
    cout << ans << endl;
    return 0;
}