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 5.4.1) (judge verdict: Runtime Error)

#include <stdio.h>
#include <string.h>
#include <stdbool.h>
#include <stdint.h>
#include <stdlib.h>
#include <limits.h>
int acs(const void *a, const void *b){return *(int*)a - *(int*)b;} /* 1,2,3,4.. */
int des(const void *a, const void *b){return *(int*)b - *(int*)a;} /* 8,7,6,5.. */
#define min(a,b) (a < b ? a: b)
#define max(a,b) (a > b ? a: b)

#define MAXN (1000)
#define MOD (998244353)


//グラフ
//#define MAXV (MAXN*MAXN)
#define MAXV (100)
int Gv;
typedef struct edge
{
    int to;
    struct edge* next;
}edge;

edge* Ge[MAXV]; //vから出てる有向辺
int Gd[MAXV]; //vの次数

// vは頂点数、指定するときは0～(v-1)
void Ginit(int v)
{
    Gv = v;
}

void Gadd(int v, int to)
{
    edge* e = malloc(sizeof(edge));
    e->to = to;
    e->next = Ge[v];
    Ge[v] = e;
    Gd[v]++;
}

// 最長経路
// 閉路があれば-1
bool GLPvisit[MAXV];
bool GLPfixed[MAXV];
int GLP[MAXV];
int GLP_DFS(int v)
{
    if(GLPvisit[v])
    {
        return GLPfixed[v] ? GLP[v] : -1;
    }
    GLPvisit[v] = true;
    GLP[v] = 0;
    edge* e = Ge[v];
    while(e!=NULL)
    {
        int l = GLP_DFS(e->to);
        if(l == -1) return -1;
        GLP[v] = max(GLP[v], l+1);
        e = e->next;
    }
    GLPfixed[v] = true;
    return GLP[v];
}
int GLongestPath(void)
{
    int lmax = 0;
    for(int i=0;i<Gv;i++)
    {
        int l = GLP_DFS(i);
        if(l==-1)
        {
            return -1;
        }
        lmax = max(lmax, l);
    }
    return lmax;
}

int sid[MAXN][MAXN];
int a[MAXN][MAXN-1];

int main(void)
{
    int n;
    scanf("%d",&n);

    int s = 0;
    for(int i=0;i<n-1;i++)for(int j=i+1;j<n;j++)
    {
        sid[i][j] = s;
        sid[j][i] = s;
        s++;
    }

    for(int i=0;i<n;i++)for(int j=0;j<n-1;j++)
    {
        scanf("%d",&(a[i][j]));
        a[i][j]--;
    }

    Ginit(s);
    for(int i=0;i<n;i++)for(int j=0;j<n-2;j++)
    {
        Gadd(sid[i][a[i][j]], sid[i][a[i][j+1]]);
    }

    int ans = GLongestPath();
    if(ans == -1)
    {
        printf("-1\n");
        return 0;
    }
    
    printf("%d\n", ans+1);
    return 0;
}