Problem: We will consider the numbers a and b as adjacent if they differ by exactly one, that is, |a-b|=1.

We will consider cells of a square matrix n × n as adjacent if they have a common side, that is, for cell (r, c) cells (r, c-1), (r, c+1), (r-1, c) and (r+1, c) are adjacent to it.

For a given number n, construct a square matrix n × n such that: 

  * Each integer from 1 to n^2 occurs in this matrix exactly once; 
  * If (r_1, c_1) and (r_2, c_2) are adjacent cells, then the numbers written in them must not be adjacent. 

Input

The first line contains one integer t (1 ≤ t ≤ 100). Then t test cases follow.

Each test case is characterized by one integer n (1 ≤ n ≤ 100).

Output

For each test case, output: 

  * -1, if the required matrix does not exist; 
  * the required matrix, otherwise (any such matrix if many of them exist). 



The matrix should be outputted as n lines, where each line contains n integers.

Example

Input


3
1
2
3


Output


1
-1
2 9 7
4 6 3
1 8 5

Solution (in CPP):

Solution:
//***********************************************************//
#include<bits/stdc++.h>
using namespace std;
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
typedef long long ll;
typedef long double ld;
#define pb push_back
#define forr(i,a,n)  for(ll i=a;i<n;i++)
#define forrr(i,a,n) for(ll i=a;i<=n;i++)
#define bp(n) __builtin_popcount(n)   // prints one's in binary representation
#define w(t)  int t;cin>>t;while(t--)
#define lc "\n"
#define men(a) (*min_element(a.begin(), a.end()))
#define mex(a) (*max_element(a.begin(), a.end()))
#define deb(x) cout << #x << " " << x << endl ;
#define dbg2(x,y) cout<<#x<<"="<<(x)<<" "<<#y<<"="<<(y)<<endl;
#define all(x) x.begin(), x.end()
#define lowb(a, x) ( lower_bound((a).begin(), (a).end(), (x)) - (a).begin())
#define uppb(a, x) ( upper_bound((a).begin(), (a).end(), (x)) - (a).begin())
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
const ll  mod =1e9+7;
const ll INF = 1e18;
const int N = 100001;
//*********************************************************//


void solve()
{
    int n , y = 0 ; 
    cin >> n  ; 
    vector<int>a(n*n+1) ;
    int j = 0 ;
    if(n == 2){
      cout << -1 << endl ;
      return ;
    }
    for(int i = 1 ; i<=n*n ; i++){
          if(i & 1 ){
            a[j] = i ;
            j++;
          }
    }
    for(int i = 1 ; i <= n*n ; i++){
      if( i % 2 == 0){
        a[j] = i ; 
        j++;
      }
    }
    y = 0 ;
    forr(i,0,j){
      cout << a[i] << " " ;
      y++;
      if( y >= n){
        y = 0 ;
        cout << endl ;
      }
    }
    




}


int main()
{

IOS;
w(t)
{solve();}
// end_routine();
return 0;
 }