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>

using namespace std;

int main() {
    int t;
    cin >> t;
    while (t--) {
        int n, k;
        cin >> n >> k;
        while (k > 3) {
            --n;
            --k;
            cout << 1 << ' ';
        }
        if (n % 2 == 1) {
            cout << n / 2 << ' ' << n / 2 << ' ' << 1 << '\n';
        }
        else if (n % 4 == 0) {
            cout << n / 2 << ' ' << n / 4 << ' ' << n / 4 << '\n';
        }   
        else {
            cout << (n - 2) / 2 << ' ' << (n - 2) / 2 << ' ' << 2 << '\n';
        }
    }
}