C. Not Adjacent Matrix

time limit per test

4 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

We will consider the numbers aa and bb as adjacent if they differ by exactly one, that is, |a−b|=1|a−b|=1.

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

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

Each integer from 11 to n2n2 occurs in this matrix exactly once;
If (r1,c1)(r1,c1) and (r2,c2)(r2,c2) are adjacent cells, then the numbers written in them must not be adjacent.

Input

The first line contains one integer tt (1≤t≤1001≤t≤100). Then tt test cases follow.

Each test case is characterized by one integer nn (1≤n≤1001≤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 nn lines, where each line contains nn integers.

Example

input

Copy

output

Copy

解题说明：此题是一道模拟题，找规律用构造法来做。

#include<stdio.h>
#include<math.h>
int main()
{long long int a[200][200], n, i, j, p, t, k;scanf("%lld", &t);while(t--){scanf("%lld", &n);if(n==2){printf("-1\n");}else{p=1;k=n+n*(n-1)/2;for(j=0; j<n-1; j++){for(i=0; i<(n-j); i++){a[i][i+j]=p;p++;if(j>0){a[i+j][i]=k;k++;}}}a[0][n-1]=n*n-1;a[n-1][0]=n*n;for(i=0; i<n; i++){for(j=0; j<n; j++){printf("%lld ", a[i][j]);}printf("\n");}}}return 0;
}

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