#include<bits/stdc++.h>
using namespace std;
typedef unsigned long long int ull;
typedef long long int ll;
typedef pair<ll,ll> pll;
typedef long double D;
//typedef complex<D> P;
#define F first
#define S second
const ll MOD=1000000007;
//const ll MOD=998244353;

template<typename T,typename U>istream & operator >> (istream &i,pair<T,U> &A){i>>A.F>>A.S; return i;}
template<typename T>istream & operator >> (istream &i,vector<T> &A){for(auto &I:A){i>>I;} return i;}
template<typename T,typename U>ostream & operator << (ostream &o,const pair<T,U> &A){o<<A.F<<" "<<A.S; return o;}
template<typename T>ostream & operator << (ostream &o,const vector<T> &A){int i=A.size(); for(auto &I:A){o<<I<<(--i?" ":"");} return o;}

//time,man
vector<vector<int>> cul(ll N){
  vector<vector<int>> A(N-1,vector<int>(N));
  if(N&1){
    for(int i=1;i<N;i++){
      for(int j=0;j<N;j++){
        A[i-1][j]=(i+j)%N;
      }
    }
    return A;
  }
  if(N==4){
    A[0]={1,3,1,0};
    A[1]={3,2,0,2};
    A[2]={2,0,3,1};
    return A;
  }
  auto B=cul(N/2);
  for(int i=0;i<N/2-1;i++){
    for(int j=0;j<N;j++){
      int idx=j%(N/2),dif=(j>=N/2?N/2:0);
      A[i][j]=B[i][idx]+dif;
    }
  }
  for(int i=N/2-1;i<N-1;i++){
    int t=i-N/2+1;
    int dif=N/2;
    for(int j=0;j<N/2;j++){
      A[i][j]=(t+j)%(N/2)+dif;
      A[i][(t+j)%(N/2)+dif]=(j+1)%(N/2);
    }
  }
  return A;
}


int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  ll N;
  cin>>N;
  if(N==2){cout<<-1<<endl; return 0;}
  auto A=cul(N);
  //for(auto &I:A){cout<<I<<endl;}
  
  for(int i=0;i<N;i++){
    for(int j=0;j+1<N;j++){cout<<A[j][i]+1<<(j+2==N?"\n":" ");}
  }
  

  return 0;
}

### Prompt

Your challenge is to write a cpp solution to the following problem:

There are N persons, conveniently numbered 1 through N. They will take 3y3s Challenge for N-1 seconds.

During the challenge, each person must look at each of the N-1 other persons for 1 seconds, in some order.

If any two persons look at each other during the challenge, the challenge ends in failure.

Find the order in which each person looks at each of the N-1 other persons, to be successful in the challenge.

Constraints

* 2 \leq N \leq 100

Input

The input is given from Standard Input in the following format:


N


Output

If there exists no way to be successful in the challenge, print `-1`.

If there exists a way to be successful in the challenge, print any such way in the following format:


A_{1,1} A_{1,2} ... A_{1, N-1}
A_{2,1} A_{2,2} ... A_{2, N-1}
:
A_{N,1} A_{N,2} ... A_{N, N-1}


where A_{i, j} is the index of the person that the person numbered i looks at during the j-th second.

Judging

The output is considered correct only if all of the following conditions are satisfied:

* 1 \leq A_{i,j} \leq N
* For each i, A_{i,1}, A_{i,2}, ... , A_{i, N-1} are pairwise distinct.
* Let X = A_{i, j}, then A_{X, j} \neq i always holds.

Examples

Input

7


Output

2 3 4 5 6 7
5 3 1 6 4 7
2 7 4 1 5 6
2 1 7 5 3 6
1 4 3 7 6 2
2 5 7 3 4 1
2 6 1 4 5 3


Input

2


Output

-1

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;
typedef unsigned long long int ull;
typedef long long int ll;
typedef pair<ll,ll> pll;
typedef long double D;
//typedef complex<D> P;
#define F first
#define S second
const ll MOD=1000000007;
//const ll MOD=998244353;

template<typename T,typename U>istream & operator >> (istream &i,pair<T,U> &A){i>>A.F>>A.S; return i;}
template<typename T>istream & operator >> (istream &i,vector<T> &A){for(auto &I:A){i>>I;} return i;}
template<typename T,typename U>ostream & operator << (ostream &o,const pair<T,U> &A){o<<A.F<<" "<<A.S; return o;}
template<typename T>ostream & operator << (ostream &o,const vector<T> &A){int i=A.size(); for(auto &I:A){o<<I<<(--i?" ":"");} return o;}

//time,man
vector<vector<int>> cul(ll N){
  vector<vector<int>> A(N-1,vector<int>(N));
  if(N&1){
    for(int i=1;i<N;i++){
      for(int j=0;j<N;j++){
        A[i-1][j]=(i+j)%N;
      }
    }
    return A;
  }
  if(N==4){
    A[0]={1,3,1,0};
    A[1]={3,2,0,2};
    A[2]={2,0,3,1};
    return A;
  }
  auto B=cul(N/2);
  for(int i=0;i<N/2-1;i++){
    for(int j=0;j<N;j++){
      int idx=j%(N/2),dif=(j>=N/2?N/2:0);
      A[i][j]=B[i][idx]+dif;
    }
  }
  for(int i=N/2-1;i<N-1;i++){
    int t=i-N/2+1;
    int dif=N/2;
    for(int j=0;j<N/2;j++){
      A[i][j]=(t+j)%(N/2)+dif;
      A[i][(t+j)%(N/2)+dif]=(j+1)%(N/2);
    }
  }
  return A;
}


int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  ll N;
  cin>>N;
  if(N==2){cout<<-1<<endl; return 0;}
  auto A=cul(N);
  //for(auto &I:A){cout<<I<<endl;}
  
  for(int i=0;i<N;i++){
    for(int j=0;j+1<N;j++){cout<<A[j][i]+1<<(j+2==N?"\n":" ");}
  }
  

  return 0;
}

```