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:
/*!enjoy karo yaar!*/

//CF,CC,AtC,SPOJ: hp1999
//HE: hemant269
//HR: hemant2132

#include<bits/stdc++.h>
using namespace std;

/*-----<Defines>-----*/
#define     int             long long int
#define     fast()          ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define     all(x)          begin(x),end(x)
#define     rz(x)           resize(x)
#define     asn(x,y)        assign(x,y)
#define     mem(a,b)        memset(a,b,sizeof(a))
#define     sz(x)           ((int)x.size())
#define     eb              emplace_back
#define     pb              push_back
#define     pf              push_front
#define     pob             pop_back
#define     pof             pop_front
#define     ins             insert
#define     vi              vector<int>
#define     pii             pair<int,int>
#define     mii             map<int,int>
#define     F               first
#define     S               second
#define     remax(a,b)      a=max(a,b)
#define     remin(a,b)      a=min(a,b)
#define     bitcount(x)     __builtin_popcountll(x)
#define     iceil(n,x)      (((n)-1)/(x)+1)
#define     flush           fflush(stdout)

using ull=unsigned long long;
using ll=long long;
using ld=long double;
/*-----</Defines>-----*/

/*-----<Debugging Stuff>-----*/
string to_string(char s)
{   string res="'''";res[1]=s;return res; }

string to_string(string s)
{   return '"'+s+'"'; }

string to_string(const char* s)
{   return to_string((string)s); }

string to_string(bool b)
{   return (b?"true":"false"); }

template<typename A, typename B>
string to_string(pair<A, B> p)
{   return "("+to_string(p.F)+", "+to_string(p.S)+")"; }

template<typename A>
string to_string(A v)
{
    bool first=1; string res="{";
    for(const auto &x:v)
    {
        if(!first)  res+=", ";
        first=0;
        res+=to_string(x);
    }
    res+="}";
    return res;
}

void debug() { cout<<"\n"; }
template<typename Head, typename... Tail>
void debug(Head H, Tail... T)
{
    cout<<" "<<to_string(H);
    debug(T...);
}

#define show(...) cout << "[" << #__VA_ARGS__ << "]:", debug(__VA_ARGS__)
/*-----</Debugging Stuff>-----*/

/*-----<Mod Operations>-----*/
const int M=1e9+7;
inline int add(int a,int b,int p=M){ a=a+b; if(a>=p) a-=p; return a; }
inline int sub(int a,int b,int p=M){ a=a-b; if(a<0) a+=p; return a; }
inline int mul(int a,int b,int p=M){ a=(a*1ll*b)%p; return a; }
/*-----</Mod Operations>-----*/

/*-----<Constants>-----*/
const ld pi=acos(-1),eps=1e-9;
const ll inf=1e18;
const int N=1;
/*-----</Constants>-----*/

void solve()
{
    int n;
    cin >> n;

    if (n == 2)
    {
        cout << "-1\n";
        return;
    }

    vector<vi> mat(n, vi(n, 0));
    int cnt = 1;
    
    for (int k=0; k<n; ++k)
    {
        for (int i=0; i<n-k; ++i)
            mat[i][i+k] = cnt++; 
    
        if (k)
        {
            for (int i=0; i<n-k; ++i)
                mat[i+k][i] = cnt++;
        }
    }

    for (int i=0; i<n; ++i)
    {
        for(int j=0; j<n; ++j)
            cout << mat[i][j] <<" ";
        cout << "\n";
    }
}

int32_t main()
{
    fast();

    int t=1;
    cin>>t;
    for(int z=1;z<=t;++z)
    {
        solve();
        // cout<<"\n";
    }

    // cerr<< "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms";

    return 0;
}