#define _CRT_SECURE_NO_WARNINGS 1
#include<bits/stdc++.h>
#include<unordered_map>
#include<unordered_set>
using namespace std;
#define ll long long
//#define double long double
#define lc u<<1
#define rc u<<1|1
#define X first
#define Y second
#define endl "\n"
#define int long long
const int N = 1e5 + 50;
const int M = 5e5 + 5;
const ll maxm = 1e18 + 5;
const ll mod = 998244353;
int dx[] = { -1,0,1,0 };
int dy[] = { 0,1,0,-1 };
//random_shuffle(id + 1, id + n + 1);
ll ksm(ll a, ll b);
ll gcd(ll a, ll b);

template<class T>inline void read(T& x) {
	x = 0;
	char c = getchar();
	while (!isdigit(c))c = getchar();
	while (isdigit(c))x = x * 10 + (c & 15), c = getchar();
}

void write(ll x)
{
	if (x < 0)
		putchar('-'), x = -x;
	if (x > 9)
		write(x / 10);
	putchar(x % 10 + '0');
	return;
}

ll mp[1005][1005];
int n, k;
int lowbit(int n) {
	return n & -n;
}

void solve() {
	cin >> n >> k;
	if (k > lowbit(n) - 1) {
		cout << "Impossible" << endl;
		return;
	}
	//不会构造，这题我根本找不到规律
	//看出来了，构造的方法应该是不断将左上矩阵复制到右下矩阵
	for (int i = 1; i <= n; i++)mp[1][i] = i;
	ll idx = 1;
	while (1) {
		bool f = 0;
		for (int j = 1; j <= n; j += idx) {//起点以idx递增
			if (f == 0) {
				for (int kk = 0; kk < idx; kk++) {
					for (int i = 1; i <= idx; i++) {
						ll p = i + idx, q = j+kk + idx;
						if (q > n)q -= n;
						mp[p][q] = mp[i][j+kk];
					}
				}
			}
			else {
				for (int kk = 0; kk < idx; kk++) {
					for (int i = 1; i <= idx; i++) {
						ll p = i + idx, q = j + kk - idx;
						if (q < 1)q += n;
						mp[p][q] = mp[i][j+kk];
					}
				}
			}
			f = !f;
		}
		idx *= 2;
		if (idx >= k+1)break;
	}
	for (int i = 2; i <= k + 1; i++) {
		cout << mp[i][1];
		for (int j = 2; j <= n; j++) {
			cout << " " << mp[i][j];
		}
		cout << endl;
	}
}


signed main()
{
	ios::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);
	int T = 1;
	cin >> T;
	//read(T);
	while (T--)
		solve();
	return 0;
}

/*PonyHex*/


ll ksm(ll a, ll b) {
	ll base = a;
	ll ans = 1;
	while (b) {
		if (b & 1)ans *= base % mod;
		ans %= mod;
		base *= base; base %= mod;
		b >>= 1;
	}
	return ans % mod;
}
ll gcd(ll a, ll b) {
	return b ? gcd(b, a % b) : a;
}