It is the hard version of the problem. The only difference is that in this version 3 ≤ k ≤ n.

You are given a positive integer n. Find k positive integers a_1, a_2, …, a_k, such that:

  * a_1 + a_2 + … + a_k = n 
  * LCM(a_1, a_2, …, a_k) ≤ n/2 



Here LCM is the [least common multiple](https://en.wikipedia.org/wiki/Least_common_multiple) of numbers a_1, a_2, …, a_k.

We can show that for given constraints the answer always exists.

Input

The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.

The only line of each test case contains two integers n, k (3 ≤ n ≤ 10^9, 3 ≤ k ≤ n).

It is guaranteed that the sum of k over all test cases does not exceed 10^5.

Output

For each test case print k positive integers a_1, a_2, …, a_k, for which all conditions are satisfied.

Example

Input


2
6 4
9 5


Output


1 2 2 1 
1 3 3 1 1 
#include <bits/stdc++.h>

#define pb push_back
#define mp make_pair
#define sz(x) (int)(x).size()
#define S second
#define F first
#define all(x) (x).begin(), (x).end()

using namespace std;
using ll = long long;

void setIO(string name = "") {
	ios_base::sync_with_stdio(NULL);
	cin.tie(NULL);
	cout.tie(NULL);

	if (sz(name)) {
		freopen((name + ".in").c_str(), "r", stdin);
		freopen((name + ".out").c_str(), "w", stdout);
	}
}

const int inf = 1e9;
const ll INF = 1e18;
const int mod = 1e9 + 7;
const int MAXN = 1e6 + 5;

ll n;
int k;

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

	vector<ll> res;

	while (sz(res) != k - 3) {
		res.pb(1);
		n--;
	}

	if (n & 1ll) {
		res.pb(1);
		res.pb((n - 1) / 2);
		res.pb((n - 1) / 2);
	}

	else if (n % 4 == 0) {
		res.pb(n / 4);
		res.pb(n / 4);
		res.pb(n / 2);
	}

	else {
		res.pb(2);
		res.pb((n - 2) / 2);
		res.pb((n - 2) / 2);
	}

	for (ll i : res) {
		cout << i << ' ';
	}

	cout << '\n';
}


int main() {
	setIO();
	
	int tt = 1;
	cin >> tt;
	while (tt--) {
		solve();
	}
	
	return 0;
}
