/*
f(a^2)=1, 
想办法 把(n-k)/2个 -1 填进去

积性函数，完全由质数位置的值决定.1e6内有78498个质数
当我们确定 sqrt(n) 以下素数的取值的时候，剩下的素数对答案的贡献是独立的

*/
#include<bits/stdc++.h>
typedef long long LL;
using namespace std;
const int N = 1e6 + 5;
const int M = 1e5 + 5;
int prime[N], lp = 0;
bool flag[N];
int f[N];
void init()
{
	flag[0] = flag[1] = true;
	for (int i = 2; i < N; i++) {
		if (!flag[i]) prime[++lp] = i;
		for (int j = 1; j <= lp && prime[j] * i < N; j++) {
			flag[prime[j] * i] = true;
			if (i % prime[j] == 0) break;
		}
	}
}
int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	init();
	int T;
	cin >> T;
	while (T--) {
		int n, k;
		cin >> n >> k;
		if (n-k&1) {
			cout << "-1\n";
		}
		else if (n >= 200) {
			for (int i = 1; i <= n; i++) f[i] = 1;
			int cnt = n-k>>1;
			int t = sqrt(n) + 1;
			int l = lower_bound(prime+1, prime+lp+1, t) - prime;
			int r = upper_bound(prime+1, prime+lp+1, n) - prime - 1;
			for (int i = l; i <= r; i++) {
				int p = n / prime[i];//-1的数量
				if (cnt >= p) {
					cnt -= p;
					for (int j = prime[i]; j <= n; j += prime[i]) f[j] = -1;
				}
			}
			for (int i = 1; i <= n; i++) cout << f[i] << ' '; cout << "\n";
//			cout << "cnt : " << cnt << "\n";
		}
		else {
			int t = sqrt(n) + 1;
			int l = lower_bound(prime+1, prime+lp+1, t) - prime;
			int r = upper_bound(prime+1, prime+lp+1, n) - prime - 1;
			t = 1 << l-1;
			for (int q = 0; q < t; q++) {//2 ^ (l-1) 次方
				for (int i = 1; i <= n; i++) f[i] = 1;
				
				int cnt = n-k>>1;
				for (int i = 1; i < l; i++) {//预处理前sqrt(n)的质数
					if (q&(1<<i-1)) {
						f[prime[i]] = -1;
						for (int j = prime[i]*2; j <= n; j += prime[i]) f[j] = f[prime[i]] * f[j/prime[i]];
					}
				}
				for (int i = 1; i <= n; i++) cnt -= (f[i] == -1);
//				cout << q << ":" << cnt << "\n";
				for (int i = l; i <= r; i++) {
					int p = 1;//-1的数量
					for (int j = prime[i]*2; j <= n; j += prime[i]) p += f[j];
					if (p > 0 && cnt - p >= 0) {
						cnt -= p;
						f[prime[i]] = -1;
						for (int j = prime[i]*2; j <= n; j += prime[i]) f[j] = (f[j]==-1?1:-1);
					}
				}
//				cout << q << " cnt:" << cnt << "\n";
				if (cnt == 0) {
					for (int i = 1; i <= n; i++) cout << f[i] << ' '; cout << "\n";
					break;
				}
			}
		}
	}
	return 0;
}